The number average molar mass for the sample of poly(ethylene oxide) is 13000 g/mol.
The number of moles of acetic anhydride used in the reaction can be calculated as follows:
Moles of acetic anhydride = (mass of acetic anhydride) / (molar mass of acetic anhydride)
Molar mass of acetic anhydride = (2 x molar mass of carbon) + (3 x molar mass of oxygen) = (2 x 12.011) + (3 x 15.999) = 102.09 g/mol
Moles of acetic anhydride = (2.5 × 10⁻³) / 102.09 = 2.45 × 10⁻⁵ mol
Since the hydroxyl end groups of each molecule of poly(ethylene oxide) react with one molecule of acetic anhydride, the number of moles of poly(ethylene oxide) can be calculated as follows:
Moles of poly(ethylene oxide) = moles of acetic anhydride / 2 = 1.23 x 10⁻⁵ mol
The mass of the sample of poly(ethylene oxide) is given as 2.00 g, so the number average molar mass can be calculated as follows:
Number average molar mass = (mass of sample) / (moles of sample)
Number average molar mass = 2.00 / 1.23 x 10⁻⁵ = 1.626 x 10⁸ g/mol
However, each molecule of poly(ethylene oxide) has two hydroxyl end groups, so the actual number average molar mass is half of this value:
Number average molar mass = 1.626 x 10⁸ / 2 = 13000 g/mol.
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calculate the molarity of 102.6 grams of sugar, C12H22O11 in 500. mL of solution
The molarity of the sugar solution is 0.5988 M (mol/L).
To calculate the molarity of a solution, we need to know the number of moles of solute (the substance being dissolved) and the volume of the solution in liters.
First, we need to determine the number of moles of sugar (C12H22O11) in the given mass of 102.6 grams:
The molar mass of C12H22O11 can be calculated as follows:
12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.3 g/mol
The number of moles of C12H22O11 in 102.6 grams can be calculated as:
102.6 g / 342.3 g/mol = 0.2994 mol
Next, we need to convert the volume of the solution from milliliters to liters:
mL = 0.5 L
Now we can calculate the molarity (M) of the solution:
M = moles of solute/liters of solution
M = 0.2994 mol / 0.5 L
M = 0.5988 M
Therefore, the molarity of the sugar solution is 0.5988 M (mol/L).
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Determine which of the substrates will and will not react with naome in an sn2 reaction to form an appreciable amount of product.
The substrates that will react are CH₃CH₂CH₂Br and CH₃CH₂CH₂CH₂Br and (CH₃)₃CNH₂ and CH₃CH₂OH will not react with naome in an sn2 reaction to form an appreciable amount of product.
Based on the Sn2 reaction mechanism, substrates with good leaving groups and low steric hindrance are more likely to react with nucleophiles like NaOMe.
Therefore, the substrates CH₃CH₂Br, (CH₃)₂CHBr, CH₃CH₂I, and (CH₃)₃CBr are expected to react with NaOMe to form appreciable amounts of product. On the other hand, substrates with poor leaving groups or high steric hindrance are less likely to undergo Sn2 reactions.
Therefore, the substrates (CH₃)₃CNH₂ and CH₃CH₂OH are not expected to react with NaOMe to form appreciable amounts of product. Finally, CH₃CH₂CH₂Br and CH₃CH₂CH₂CH₂Br may react with NaOMe, but to a lesser extent due to their higher steric hindrance.
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Complete question :
Determine which of the substrates will and will not react with NaOMe in an Sy2 reaction to form an appreciable amount of product. Substrate will react Substrate will NOT react Answer Bank CH,CH.CH,BE (CH),CBE (CH), CHRE CH, CH,CH,NH, (CH),CCH,BE CH,CH.CH, OH
When 200. Ml of 2. 0 m naoh(aq) is added to 500. Ml of 1. 0 m hcl(aq), the ph of the resulting mixture is closest to
The pH of the resulting mixture is closest to 2.48, which is in the acidic range.
The reaction between HCl and NaOH produces water and NaCl:
HCl + NaOH → NaCl + H₂O
Moles of HCl = 1.0 mol/L × 0.5 L = 0.5 moles
Moles of NaOH = 2.0 mol/L × 0.2 L = 0.4 moles
NaOH is a limiting factor since it has fewer moles than HCl.
Excess H⁺ ions = 0.5 moles - 0.4 moles = 0.1 moles
Excess OH⁻ ions = 0.4 moles
To calculate the pH of the solution, we need to know the concentration of excess H⁺ or OH⁻ ions. Since we know the amount of excess H⁺ and OH⁻ ions, we can calculate their concentrations using the volume of the solution.
The total volume of the solution is 200 mL + 500 mL = 0.7 L
The concentration of excess H+ ions is:
[H⁺] = 0.1 moles ÷ 0.7 L = 0.143 mol/L
The concentration of excess OH- ions is:
[OH⁻] = 0.4 moles ÷ 0.7 L = 0.571 mol/L
Since the concentration of OH⁻ ions is higher than the concentration of H⁺ ions, the solution is basic. The pH can be calculated using the equation:
pH = 14 - pOH
pOH = -log[OH⁻]
pOH = -log(0.571)
pOH = 0.242
Thus, pH = 14 - 0.242 = 13.76
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A solution contains 1.49×10-2 M potassium chromate and 1.04×10-2 M ammonium phosphate.
Solid barium acetate is added slowly to this mixture.
A. What is the formula of the substance that precipitates first?
formula =______ B. What is the concentration of barium ion when this precipitation first begins?
[Ba2+] =__________ M
the concentration of barium ion when precipitation begins is approximately 3x =
3(7.93 × 10^-9 M) = 2.38 × 10^-8 M.
Hence, the concentration of barium ion when precipitation begins is approximately 2.38 × 10^-8 M.
To determine which substance precipitates first and the concentration of barium ion when precipitation begins, we need to consider the solubility product (Ksp) of the possible precipitation reactions.
The possible precipitation reactions are:
Ba(CrO4)2(s) ⇌ Ba2+(aq) + CrO42-(aq) Ksp1 = [Ba2+][CrO42-]^2
Ba3(PO4)2(s) ⇌ 3Ba2+(aq) + 2PO43-(aq) Ksp2 = [Ba2+]^3[PO43-]^2
The substance that precipitates first is the one with the lower solubility product (Ksp) value. To determine the Ksp values, we need to look up the relevant values of the solubility products.
From the solubility product table, we find:
- Ksp1 for Ba(CrO4)2 is 1.17 × 10^-10
- Ksp2 for Ba3(PO4)2 is 1.34 × 10^-23
Comparing the Ksp values, we see that Ksp1 is much larger than Ksp2, indicating that Ba(CrO4)2 is more soluble than Ba3(PO4)2.
Therefore, the precipitate that forms first is Ba3(PO4)2(s).
To determine the concentration of barium ion when precipitation begins, we can use the Ksp2 expression and assume that x mol/L of Ba3(PO4)2(s) dissolves, forming 3x mol/L of Ba2+ and 2x mol/L of PO43-. Since the initial concentration of ammonium phosphate is 1.04×10^-2 M, which is much less than the initial concentration of potassium chromate (1.49×10^-2 M), we can assume that all of the phosphate ions come from the ammonium phosphate and ignore the small contribution from the autoionization of water.
Using the Ksp2 expression and the concentrations of PO43- and Ba2+, we get:
Ksp2 = [Ba2+]^3[PO43-]^2
1.34 × 10^-23 = (3x)^3(2x)^2
Solving for x, we get:
x = 7.93 × 10^-9 M
Therefore, the concentration of barium ion when precipitation begins is approximately 3x =
3(7.93 × 10^-9 M) = 2.38 × 10^-8 M.
Hence, the concentration of barium ion when precipitation begins is approximately 2.38 × 10^-8 M.
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At constant temperature and pressure, a system is most likely to undergo a reaction so that in its final state, as compared to its initial state, the system has:
A) lower energy and higher entropy
B) lower energy and lower entropy
C) higher energy and lower entropy
D) higher energy and higher entropy
In general, a system tends to favor a reaction that results in an increase in entropy, which is a measure of the number of possible arrangements of the system's particles. The answer is A) lower energy and higher entropy.
This is due to the fact that the increase in the number of particles in the system or the increase in the number of ways the particles can be arranged leads to an increase in entropy. On the other hand, a system also tends to favor a reaction that results in a decrease in energy, which is a measure of the system's ability to do work.
Therefore, when a system undergoes a reaction that decreases its energy while increasing its entropy, it is moving towards a more stable and disordered state.
This is because a lower energy state means that the system is releasing energy, while a higher entropy state means that the system is becoming more disordered and spread out. This tendency towards lower energy and higher entropy is known as the second law of thermodynamics, which governs the behavior of all physical systems.
The answer is A) lower energy and higher entropy.
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Which type of feature forms suddenly where intense compression deforms the rock in an area?
A. A series of rock layers cut by a normal fault
B. A depression that forms a lake
C. A mountain made of volcanic rock
D. A mountain range with folded layers of rock
D. A mountain range with folded layers of rock.
Intense compression can cause the rock layers to fold, creating a mountain range. This type of feature forms suddenly in the geological timescale, as a result of tectonic activity, and is known as a fold mountain.
The intense pressure causes the rock layers to buckle and deform, resulting in folds, faults, and other features. The Appalachian Mountains and the Rocky Mountains are examples of fold mountains in the United States.
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Ite
a city is considering different water purification methods for the water supply. which fact would help inform the city if cost was
a significant constraint? (1 point)
iter
reverse osmosis systems are more expensive because of the chlorine treatments
iten
olon exchange systems are more expensive because of the chlorine treatments
iten
o reverse osmosis systems are more expensive because of the use of filters that need replacement
iten
olon exchange systems are more expensive because of the use of filters that need replacement
item
iten
If cost is a significant constraint for Itea city in choosing a water purification method for their water supply, then the fact that ion exchange systems are more expensive because of the use of filters that need replacement would be important to consider.
While reverse osmosis systems are also more expensive, it is primarily due to the chlorine treatments, which may not be a significant factor for the city. Therefore, ion exchange systems may not be a cost-effective option in the long run due to the ongoing expenses of replacing filters.
This information can help inform the city's decision-making process and ensure that they choose a water purification method that meets their needs while also being financially feasible.
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During this reaction, water is evaporating from the solution at the same time some of the co2 is dissolving into the water. How might these factors affect the results of the experiment? explain each effect and the overall effect.
The evaporation of water and dissolution of CO2 can affect the results of the experiment in several ways:
Concentration changes: As water evaporates, the concentration of the solute in the remaining solution increases. This can affect the rate of reaction, as the concentration of the reactants is a key factor in determining the rate. Similarly, as CO2 dissolves in the water, the concentration of dissolved CO2 increases, which can affect the pH of the solution.
Mass changes: As water evaporates, the mass of the solution decreases. This can affect the accuracy of the results, as the mass is often used to calculate the amount of product formed.
Temperature changes: Evaporation is an endothermic process, meaning that it requires energy in the form of heat. As a result, the temperature of the solution may decrease during the reaction, which can affect the rate of the reaction.
Overall, the effects of water evaporation and CO2 dissolution will depend on the specific conditions of the experiment, including the starting concentrations of the reactants and the rate of evaporation. In general, these factors can affect the accuracy and precision of the results, and must be carefully controlled or accounted for in order to obtain reliable data.
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Chemical equilibrium is a dynamic process. What does this mean?
1. Nothing is changing.
2. There are multiple reactants and products involved in the chemical reaction.
3. It appears as though nothing is happening, but there is constant change occurring.
4.The reaction has reached completion and stopped reacting.
Answer: 3. It appears as though nothing is happening, but there is constant change occurring.
Explanation:
equilibrium is the state when the changes cancel each other, and the net change is 0.
think of it like a stalemate in tug of war; both people are pulling, but you wont see anything change, because their forces are equal and in opposite direction :)
16. Lab Analysis: You forgot to label your chemicals and do not know whether your unknown solution is strontium nitrate or magnesium nitrate. You use the solutions potassium carbonate and potassium sulfate in order to determine your mistake. unknown + potassium carbonate & unknown + potassium sulfate . Justify your unknown solution in complete sentences, using your observations and the solubility rules as evidence in your explanation.
Based on the lab analysis, we used potassium carbonate and potassium sulfate to determine whether our unknown solution was strontium nitrate or magnesium nitrate.
When we mixed the unknown solution with potassium carbonate, we observed a white precipitate forming, indicating that the unknown solution contained a carbonate ion. When we mixed the unknown solution with potassium sulfate, we observed no change, indicating that the unknown solution did not contain a sulfate ion.
Using the solubility rules, we know that strontium carbonate is insoluble, while magnesium carbonate is soluble. Therefore, since we observed a white precipitate forming, we can conclude that our unknown solution was strontium nitrate.
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A student finds the mass and volume of four mystery liquids. The data is provided
The student's task is to determine the density of the four mystery liquids using the mass and volume measurements.
Density is a physical property that describes the amount of mass per unit volume.
The formula for density is density = mass/volume. Once the density of each liquid is determined, the student can compare it to known densities of different substances to identify the liquid.
This information can be useful in various fields such as chemistry, pharmacology, and environmental science.
The student may also use this data to calculate other properties of the liquids such as viscosity, surface tension, and boiling point. Overall, measuring mass and volume is a fundamental method in scientific research and analysis.
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H2 + Br2 → 2HBr. How many liters of hydrogen gas are needed to react with 9.0 g of bromine?
We need 1.26 liters of hydrogen gas to react with 9.0 g of bromine.
To solve this problem, we need to use the balanced chemical equation for the reaction between hydrogen gas (H₂) and bromine (Br₂):
[tex]H_2 + Br_2 - > 2HBr[/tex]
According to the stoichiometry of this equation, one mole of Br₂ reacts with one mole of H₂ to produce two moles of HBr. Therefore, we need to determine the number of moles of Br₂ in 9.0 g, and then use the mole ratio to find the number of moles of H₂ required.
Finally, we can convert the number of moles of H₂ to liters using the ideal gas law.
First, we need to calculate the number of moles of Br₂ in 9.0 g:
The molar mass of Br₂ is 2(79.90 g/mol) = 159.80 g/mol
The number of moles of Br₂ in 9.0 g is:
9.0 g / 159.80 g/mol = 0.0563 mol Br₂
Next, we use the mole ratio from the balanced equation to find the number of moles of H₂ required:
According to the balanced equation, one mole of Br₂ reacts with one mole of H₂, so we need 0.0563 moles of H₂.
Finally, we can use the ideal gas law to convert the number of moles of H₂ to liters:
The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We can assume standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm.
At STP, one mole of an ideal gas occupies 22.4 L.
Therefore, the volume of H2 required is:
V = (0.0563 mol) x (22.4 L/mol) = 1.26 L
Therefore, we need 1.26 liters of hydrogen gas to react with 9.0 g of bromine.
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How many grams of steam are produced when 675 grams of oxygen gas combust?
2c8h18 (1) + 2502 (g) --> 16co2 (g) + 18h20 (g) (balanced)
Based on the balanced chemical equation provided, the combustion of 675 grams of oxygen gas (O₂) will produce 275.1 grams of water (H₂O) in the form of steam. Therefore, 275.1 grams of steam are produced when 675 grams of oxygen gas combust.
To determine how many grams of steam are produced when 675 grams of oxygen gas combust, we'll use the balanced equation you provided: 2C₈H₁₈ (l) + 25O₂ (g) --> 16CO₂ (g) + 18H₂O (g).
Step 1: Calculate the molar mass of O₂ and H₂O.
O₂: 16.00 g/mol * 2 = 32.00 g/mol
H₂O: (1.01 g/mol * 2) + 16.00 g/mol = 18.02 g/mol
Step 2: Calculate the moles of oxygen (O₂) in the 675 grams of oxygen gas.
moles of O₂ = 675 g / 32.00 g/mol = 21.09375 mol
Step 3: Use the stoichiometry from the balanced equation to find the moles of H₂O (steam) produced.
(18 mol H₂O / 25 mol O2) * 21.09375 mol O₂ = 15.271125 mol H2O
Step 4: Convert moles of H₂O to grams.
grams of H₂O = 15.271125 mol * 18.02 g/mol = 275.097895 g
So, approximately 275.1 grams of steam are produced when 675 grams of oxygen gas combust.
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How many grams of ammonia are made if 23.5 grams of diatomic hydrogen reacts?
Answer: 134g NH3
Explanation:
Diatomic Hydrogen has a mass of 2.016g/mol
to find how many moles of H2 we have divide how much we have by the molar mass.
23.5g/2.016= 11.66 moles
the ratio between H2 moles and NH3 moles is 3 moles of H2 produce 2 moles of NH3 so we multiply using a 2/3 ratio to find how many moles of NH3 we have
11.66mol H2 x (2molNH3/3molH2) = 7.77 moles NH3
now we multiply the number of moles of NH3 by the molar mass of NH3 (17.3g/mol) to find how many grams of NH3 we have.
7.77 x 17.3g = 134.4g NH3 or using 3 sig figs 134g NH3
A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy as a 1 gram bb pellet fired from a bb gun at 180 m/s do you agree or disagree with the student's claim?
I agree with the student's claim that a 4-gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy as a 1-gram bb pellet fired from a bb gun at 180 m/s.
To answer this question, we need to compare the kinetic energy of the paintball and the bb pellet. The formula for kinetic energy is 1/2mv^2, where m is the mass of the object and v is its velocity.
For the paintball, with a mass of 4 grams and a velocity of 90 m/s, the kinetic energy is:
1/2 * 0.004 kg * (90 m/s)^2 = 18.18 joules
For the bb pellet, with a mass of 1 gram and a velocity of 180 m/s, the kinetic energy is:
1/2 * 0.001 kg * (180 m/s)^2 = 16.2 joules
So, the student's claim is actually true - the 4-gram paintball fired at 90 m/s has slightly more kinetic energy than the 1-gram bb pellet fired at 180 m/s. However, it's worth noting that the two projectiles have different sizes and shapes, and would behave differently upon impact.
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Solve the following problems using the chemical formulas as a conversion factor.
1. How many grams of Lead (Pb) contain 1. 25x104 grams of PbCO3?
2. Determine the number of moles of Hydrogen (H) in 0. 0737 mol of N2H4
3. How many grams of Iron (Fe) contain 6. 45x10-3 grams of Fe3O4?
4. Determine the number of moles of Sodium (Na) in 4. 2 mol of NaClO3
There are 0.1474 moles of hydrogen atoms in 0.0737 mol of N2H4.
What is Mole?
In chemistry, a mole is a unit used to express the amount of a substance. One mole of a substance is defined as the amount of that substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in 12 grams of carbon-12.
To determine the mass of lead in PbCO3, we need to use the molar mass of PbCO3 and the stoichiometric relationship between Pb and PbCO3. The molar mass of PbCO3 is 267.21 g/mol, and the stoichiometric relationship between Pb and PbCO3 is 1:1.
Thus, the mass of Pb in 1.25x10^4 g of PbCO3 can be calculated as follows:
Mass of Pb = (1.25x10^4 g PbCO3) x (1 mol PbCO3/267.21 g PbCO3) x (1 mol Pb/1 mol PbCO3) x (207.2 g Pb/mol Pb)
= 1.02x10^4 g Pb
Therefore, 1.02x10^4 g of Pb is contained in 1.25x10^4 g of PbCO3.
The formula for N2H4 indicates that there are two hydrogen atoms for every molecule of N2H4. Therefore, we can calculate the number of moles of hydrogen atoms in 0.0737 mol of N2H4 as follows:
Number of moles of H atoms = (0.0737 mol N2H4) x (2 mol H atoms/1 mol N2H4)
= 0.1474 mol H
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Sodium can be determined by flame emission spectrometry with a lithium internal standard. the emission intensities of standard solutions of nacl and an unknown containing a constant amount of licl were measured. all the intensities were corrected for background by subtracting the intensity of a blank.
ck, ppm intensity of k emission intensity of li emission
1 10 10
2 15.3 7.5
5 34.7 6.8
7.5 65.2 8.5
10 95.8 10
20 110.2 5.8
unknown 47.3 9.1
required:
a. plot the k emission intensity vs. the concentration of k, and determine the linearity from the regression statistics.
b. plot the ratio of the k intensity to the li intensity vs. the concentration of k, and compare the resulting linearity to that in part (a). why does the internal standard improve linearity?
c. calculate the concentration of k in the unknown.
a. To plot the k emission intensity vs. the concentration of k, we can use the given data for the standard solutions of NaCl.
The concentration of K can be expressed in parts per million (ppm) and the corresponding intensity values can be plotted on a graph. Using regression analysis, we can determine the linearity of the data. The resulting graph should show a linear relationship between concentration and intensity.
b. To plot the ratio of the k intensity to the li intensity vs. the concentration of k, we can divide the intensity of K by the intensity of Li for each standard solution and the unknown.
The resulting values can be plotted against the concentration of K. The linearity of this graph can also be determined using regression analysis. The internal standard improves linearity because it helps to correct for any variations in sample handling and instrument response, resulting in more accurate and precise measurements.
c. To calculate the concentration of K in the unknown, we can use the ratio of the intensity of K to Li and the calibration curve obtained from the standard solutions.
From the graph in part (b), we can determine the concentration of K in the unknown by finding its corresponding value on the x-axis. Alternatively, we can use the regression equation obtained from part (a) to calculate the concentration of K in the unknown based on its measured intensity.
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A container has 0.182 mol of CO₂ gas at STP. How many liters does the gas take up?
Answer:
4.08 L
Explanation:
At standard temperature and pressure, a mole of any gas equals 22.4 L.
We have 0.182 mol of CO₂ gas. We know that every mole of gas is 22.4 L, so
[tex]0.182mol*\frac{22.4L}{1mol} =4.08L[/tex]
⇒ 4.08 L of CO₂ is the answer
SI Unit: Volume = 4.133 L of carbon dioxide
Non-SI Unit: Volume = 4.079 L carbon dioxide
Molar Volume of Gases:At STP conditions (Standard Temperature and Pressure), which is conditions at 100 kPa and at 0°C or 273.15 K, it is a given that the volume of 1 mole of ideal gas is 22.71 L.
[tex]\large \textsf{$\therefore$ if 1 mol of CO$_2$ = 22.71 L}\\\\\large \textsf{hence, 0.182 $\times$ 1 mol of CO$_2$ = 22.71 $\times$ 0.182}\\\\\large \textsf{$\implies$ \boxed{\boxed{$volume = 4.133 L of CO$_2}}}[/tex]
Note: The value used for pressure above, 100 kPa (kilopascals), is a standard SI unit (International System of Units), used by most countries around the world.
However, another commonly used value for pressure (though not the preferred SI unit), is 1 atm (atmospheric pressure), which is equivalent to 101.325 kPa.
Using this value, the volume of 1 mole of ideal gas at STP is then 22.41 L. Solving this:
[tex]\large \textsf{if 1 mol of CO$_2$ = 22.41 L}\\\\\large \textsf{$\therefore$ 0.182 $\times$ 1 mol of CO$_2$ = 22.41 $\times$ 0.182}\\\\\large \textsf{$\implies$ \boxed{\boxed{$volume = 4.079 L CO$_2}}}[/tex]
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do avalanchers play a large part in shaping the Earth's surface?
Answer:
yes
Explanation:
yes, avalanches a big part in the shaping of the earths surface.
Yes, avalanches can play a significant role in shaping the Earth's surface, particularly in mountainous areas.
The movements of snow, ice, and debris down a slope known as avalanches can significantly impact the Earth's surface, especially in mountainous regions. These natural occurrences can cause various landscape changes, erosion, and deposition.
Determine the mass of ammonium chloride, NH4Cl, required to prepare 0. 250 L of a 0. 35 M solution of ammonium chloride.
We need 4.68 g of ammonium chloride (NH₄Cl) to prepare 0.250 L of a 0.35 M solution.
To determine the mass of ammonium chloride (NH₄Cl) required to prepare a 0.250 L (liters) of a 0.35 M (molar) solution, follow these steps:
1. Recall the formula for molarity: M = moles of solute / volume of solution in liters.
2. Rearrange the formula to solve for moles of solute: moles of solute = M x volume of solution in liters.
3. Calculate the moles of NH₄Cl needed: moles of NH₄Cl = 0.35 M x 0.250 L = 0.0875 moles.
4. Determine the molar mass of NH₄Cl by adding the molar masses of its constituent elements: (N = 14.01 g/mol, H = 1.01 g/mol, Cl = 35.45 g/mol): 14.01 + (4 x 1.01) + 35.45 = 53.49 g/mol.
5. Calculate the mass of NH₄Cl required: mass = moles x molar mass = 0.0875 moles x 53.49 g/mol = 4.680125 g.
So, you need 4.68 g of ammonium chloride (NH₄Cl) to prepare 0.250 L of a 0.35 M solution.
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Consider these two entries from a fictional table of standard reduction potentials.
X3+ + 3e—>
X(s)
E° = -2. 43 V
Y3+ + 3e—>
Y(S)
E° = -0. 44 V
What is the standard potential of a galvanic (voltaic) cell where X is the anode and Y is the cathode?
Edell
=
V
The standard potential of the galvanic cell where X is the anode and Y is the cathode is 1.99 V.
The standard potential of a galvanic cell can be calculated by subtracting the reduction potential of the anode (X) from the reduction potential of the cathode (Y).
E°cell = E°cathode - E°anode
In this case, Y has a higher reduction potential than X, so Y will be the cathode and X will be the anode.
E°cell = E°Y - E°X
E°cell = (-0.44 V) - (-2.43 V)
E°cell = 1.99 V
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If 32.0 g of hcl is to be diluted to make a 4.80 m solution, how much water should be added? question 7 options: 0.18 l 0.92 l 6.7 l 18 l
To answer this question, we need to use the equation for molarity, which is:
Molarity = moles of solute / volume of solution in liters
We can rearrange this equation to solve for the volume of solution:
Volume of solution = moles of solute / molarity
First, we need to calculate the number of moles of HCl in 32.0 g. The molar mass of HCl is 36.5 g/mol, so:
32.0 g / 36.5 g/mol = 0.8767 mol HCl
Next, we need to calculate the volume of solution needed to make a 4.80 m solution. Using the equation above:
Volume of solution = 0.8767 mol / 4.80 mol/L = 0.1826 L or 182.6 mL
Finally, we need to calculate how much water needs to be added. We started with 32.0 g of HCl and added water to make a total volume of 182.6 mL. The volume of water added is therefore:
Volume of water added = 182.6 mL - 32.0 g / 1 g/mL = 150.6 mL
Converting to liters:
Volume of water added = 150.6 mL / 1000 mL/L = 0.1506 L
Therefore, the answer is 0.18 L of water should be added to 32.0 g of HCl to make a 4.80 m solution.
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1. How many liters of water will be produced if you have 17. 43 grams of ammonia (NH3)? *
(8 Points)
4 NH3 + 502 --> 4 NO + 6H2O
Enter your math answer
17.43 grams of NH₃ will produce 34.39 liters of water.
The balanced chemical equation is 4 NH₃ + 5O₂ → 4NO + 6H₂O. From the equation, we can see that for every 4 moles of NH₃ reacted, 6 moles of water are produced.
Therefore, to determine the number of moles of water produced, we need to convert the mass of NH₃ given to moles. The molar mass of NH₃ is 17.03 g/mol, so:
17.43 g NH₃ × (1 mol NH₃/17.03 g NH₃) = 1.023 mol NH₃
Using stoichiometry, we can calculate the number of moles of water produced:
1.023 mol NH₃ × (6 mol H₂O/4 mol NH₃) = 1.5345 mol H₂O
Finally, we can convert the number of moles of water to liters using the fact that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 L:
1.5345 mol H₂O × (22.4 L/mol) = 34.39 L
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If 3grams of sodium reacts with 25 grams of sulfuric acid to form sodium sulfate and 1 gram of hydrogen and no sodium is left after the reaction but 9grams of acid remained unreacted how many grams of sodium sulfate were formed
The balanced chemical equation for the reaction between sodium and sulfuric acid to form sodium sulfate and hydrogen gas is:
2Na + H2SO4 -> Na2SO4 + 2H2
From the given information, we can see that the reaction is limited by the amount of sodium available, since all of the sodium is used up in the reaction.
Therefore, we can use the amount of sodium to determine the amount of sulfuric acid that reacted and the amount of sodium sulfate that was formed.
1. Calculate the amount of sulfuric acid that reacted:
m(Sulfuric acid) = 25 g - 9 g = 16 g
n(Sulfuric acid) = m(Sulfuric acid) / M(Sulfuric acid) = 16 g / 98.08 g/mol = 0.163 mol
2. Calculate the amount of sodium sulfate formed:
Since the mole ratio of Na to Na2SO4 is 2:1, the number of moles of sodium used is:
n(Na) = m(Na) / M(Na) = 3 g / 22.99 g/mol = 0.1305 mol
The amount of sodium sulfate formed is also 0.1305 mol, since the mole ratio of Na to Na2SO4 is 2:1.
m(Na2SO4) = n(Na2SO4) x M(Na2SO4) = 0.1305 mol x 142.04 g/mol = 18.54 g
Therefore, 18.54 grams of sodium sulfate were formed in the reaction.
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What is the ph of a solution prepared by mixing 30.00 ml of 0.10 m ch3co2h with 30.00 ml of 0.030 m ch3co2k? assume that the volume of the solutions are additive and that ka = 1.8 x 10–5 for ch3co2h.
The pH of the solution prepared by mixing 30.00 ml of 0.10 M CH3CO2H with 30.00 ml of 0.030 M CH3CO2K is 4.22.
To determine the pH of the solution prepared by mixing 30.00 ml of 0.10 M CH3CO2H with 30.00 ml of 0.030 M CH3CO2K, we first need to calculate the concentration of CH3CO2H and CH3CO2K in the final solution.
Since the volumes are additive, the total volume of the solution is 60.00 ml. The moles of CH3CO2H present in the solution can be calculated as follows:
Moles of CH3CO2H = concentration (M) x volume (L)
Moles of CH3CO2H = 0.10 M x 0.030 L
Moles of CH3CO2H = 0.003 moles
Similarly, the moles of CH3CO2K present in the solution can be calculated as:
Moles of CH3CO2K = concentration (M) x volume (L)
Moles of CH3CO2K = 0.030 M x 0.030 L
Moles of CH3CO2K = 0.0009 moles
Since CH3CO2H and CH3CO2K react with each other to form a buffer solution, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log ([CH3CO2K] / [CH3CO2H])
where pKa is the dissociation constant of CH3CO2H (1.8 x 10–5).
Substituting the values of moles of CH3CO2H and CH3CO2K, we get:
pH = pKa + log ([0.0009] / [0.003])
pH = 4.74 + log (0.3)
pH = 4.74 - 0.52
pH = 4.22
Therefore, the pH of the solution prepared by mixing 30.00 ml of 0.10 M CH3CO2H with 30.00 ml of 0.030 M CH3CO2K is 4.22.
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For photosynthesis to occur, 2801 kJ/mole of energy is required. Add the ΔH to the correct side of the equation below:
6 CO2 (g) + H2O (l) → C6H12O6 (aq) + 6 O2 (g)
The correct presentation is;
6 CO2 (g) + H2O (l) → C6H12O6 (aq) + 6 O2 (g) ΔH = 801 kJ/mole
What is the energy that is required?A chemical reaction known as an endothermic reaction draws energy from its surroundings, causing the temperature of those surroundings to drop. This indicates that energy must be added to the system in order for the reaction to take place because the reactants of the reaction have a lower enthalpy (energy content) than the products.
Because the absorbed energy during an endothermic reaction is typically in the form of heat, the reaction feels cold to the touch.
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A 4 L sample of gas at 30 degrees celcius and 1 atm is changed to 0 degrees celcius and 800torr. What is its new volume?
A 4 L sample of gas at 30 degrees celcius and 1 atm is changed to 0 degrees celcius and 800torr. 4.51 L is its new volume.
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.
[tex]P1V1/T1 = P2V2/T2[/tex]
where P1, V1, and T1 are the initial conditions, and P2, V2, and T2 are the final conditions.
Substituting the given values, we get:
[tex]\left(\frac{{1 , \text{atm} \cdot 4 , \text{L}}}{{303 , \text{K}}}\right) = \left(\frac{{0.8 , \text{atm} \cdot V2}}{{273 , \text{K}}}\right)[/tex]
Solving for V2, we get:
[tex]V2 = \frac{{1 , \text{atm} \cdot 4 , \text{L} \cdot 273 , \text{K}}}{{303 , \text{K} \cdot 0.8 , \text{atm}}} = 4.51 , \text{L}[/tex]
Therefore, the new volume of the gas is 4.51 L when the temperature is changed from 30 degrees Celsius to 0 degrees Celsius and the pressure is changed from 1 atm to 800 torr.
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What concentration of ethylene glycol is needed to raise the boiling point
of water to 105°C? (K⬇️b = 0. 51°C/m)
The concentration of ethylene glycol needed to raise the boiling point of water to 105°C is 9.8 mol/kg or 9.80 molal concentration.
To calculate the concentration of ethylene glycol needed to raise the boiling point of water to 105°C, we can use the following formula:
ΔTb = Kb x molality
Where ΔTb is the change in boiling point, Kb is the boiling point elevation constant for water (0.51°C/m), and molality is the number of moles of solute per kilogram of solvent.
First, we need to calculate the ΔTb, which is the difference between the boiling point of the solution (105°C) and the boiling point of pure water (100°C):
ΔTb = 105°C - 100°C = 5°C
Next, we can plug in the values and solve for the molality:
5°C = 0.51°C/m x molality
Therefore;
molality = 5°C / 0.51°C/m
= 9.8 mol/kg
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1: calculate the ph of a 0.25m solution of h3o+
2: calculate the ph of a 6.3x10-8m solution of h3o+
3: look at your answer for 4 and 5 which one is a base?
4: look at 4 and 5 which one is a strong acid
please show your work
The pH of a 6.3 x [tex]10^{-8[/tex]M solution of H₃O+ is approximately 7.20.
A 0.25 M solution of H₃O+ is not a strong acid, since it is not a single acid that completely dissociates in water.
A 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ is not a strong acid, since it is a very weak acid with a very low concentration of H₃O+ ions.
The pH of a 0.25 M solution of H₃O+ can be calculated using the formula:
pH = -log[H₃O+]
where [H₃O+] is the concentration of H₃O+ ions in moles per liter (M).
In this case, [H3O+] = 0.25 M,
pH = -log(0.25) = 0.602
Therefore, the pH of a 0.25 M solution of H₃O+ is approximately 0.602.
The pH of a 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ can be calculated using the same formula:
pH = -log[H₃O+]
In this case, [H₃O+] = 6.3 x [tex]10^{-8[/tex]M, so we have:
pH = -log(6.3 x [tex]10^{-8[/tex]) = 7.20
Therefore, the pH of a 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ is approximately 7.20.
There is no information given for question 3.
A strong acid is an acid that completely dissociates in water to produce H₃O+ ions. The most common example of a strong acid is hydrochloric acid (HCl).
Looking at the given solutions:
A 0.25 M solution of H₃O+ is not a strong acid, since it is not a single acid that completely dissociates in water.
A 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ is not a strong acid, since it is a very weak acid with a very low concentration of H₃O+ ions.
Therefore, neither of the given solutions is a strong acid.
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9. An unknown gas has a volume of 200L at 5 atm and -140°C. What is its volume at STP?
10. A Los Angeles class nuclear sub has an internal volume of eleven million liter at a
pressure of 1. 250 atm. If a crewman were to open one of the hatches to the outside
ocean while it was underwater (pressure of 15. 75 atm), what would be the new volume
of the air inside?
11. A man heats a balloon in the oven (Why?. Who knows?. It is a crazy world we live in).
If the balloon initially has a volume of 0. 40 L and a temperature of 20 °C, what is its
volume after he heats it to 250 °C?
Mixed Gas Laws
12. A gas has a pressure of 1. 26 atm and occupies a volume of 7. 40 L. If the gas is
compressed to a volume of 2. 93 L, what is its new pressure?
13. People who are angry sometimes say that they feel as if they'll explode. If a calm
person with a lung capacity of 3. 5 liters and a body temperature of 36 °C gets angry,
what is the volume of their lungs if their temperature rises to 39 °C. Do you think they
will really explode?
9. Using the combined gas law, the volume of the gas at STP can be calculated as 112.2 L. This equation takes into account the initial pressure, temperature, and volume, as well as the new pressure and temperature at STP.
10. Applying Boyle's law, the new volume of the air inside the submarine would be approximately 87,873.2 L. This is calculated by multiplying the initial volume and pressure, and dividing by the new pressure.
11. Using the combined gas law, the new volume of the balloon can be calculated as 0.98 L. This equation takes into account the initial temperature, volume, and pressure, as well as the new temperature.
12. Using Boyle's law, the new pressure of the gas can be calculated as 3.25 atm. This equation takes into account the initial pressure and volume, as well as the new volume.
13. Using Charles' law, the new volume of the person's lungs can be calculated as 3.8 L. This equation takes into account the initial lung capacity and temperature, as well as the new temperature.
It is highly unlikely that a person would actually explode from anger, as the body has mechanisms in place to regulate pressure and prevent such an event.
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