The pair of elements that have the most similar Lewis structures are F and Ne.
The Lewis structure of an atom shows its valence electrons as dots around the symbol representing the nucleus. Atoms with similar Lewis structures have similar numbers of valence electrons and, therefore, similar chemical properties.
N and S have different numbers of valence electrons (5 and 6, respectively), so their Lewis structures are different.
P and F have different numbers of valence electrons (5 and 7, respectively), and the placement of the dots on their Lewis structures is different.
F and Ne both have 7 valence electrons and would have similar Lewis structures, with one dot representing each valence electron. They are both nonmetals and are found in the same period of the periodic table, which also contributes to their similar chemical properties.
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9. Arrange the following ions in terms of increasing atomic radius (arrange then increasing from left [smallest] to right [largest]): Ca2+, K+, Rb+, Sr2+, Na+
The ions arranged in terms of increasing atomic radius from left to right are: Ca²⁺, Sr²⁺, Na⁺, K⁺, Rb⁺.
As we move from left to right across the periodic table, due to the increasing nuclear charge the number of protons in the nucleus increases, pulling the electrons closer to the center and decreasing the atomic radius. However, as you move down a group, the number of electron shells increases, which increases the distance between the nucleus and outermost electrons, increasing the atomic radius.
Cations (positively charged ions) have smaller radii than their corresponding neutral atoms due to the loss of electrons and increased effective nuclear charge. Ca²⁺, Sr²⁺ have a +2 charge and; K⁺, Rb⁺, and Na⁺ have a +1 charge. Higher charge leads to a smaller atomic radius.
Ca²⁺, Sr²⁺ are located in Group 2, while K⁺, Rb⁺, and Na⁺ are located in Group 1 of periodic table. Arrange the ions based on their positions in the periodic table and their charges.
Based on these factors, the correct order of ions in terms of increasing atomic radius is: Ca²⁺ (smallest), Sr²⁺, Na⁺, K⁺, and Rb⁺ (largest).
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Of the following compounds, which is the most ionic? A) SiCl4 B) BrCl C)PCl3 D) Cl2O E) CaCl
A 3. 245g sample of titanium chloride was reacted with sodium metal, producing sodium chloride and metallic titianium. After the sodium chloride was washed out, the remaining titanium metal weighed 0. 819g. What is the empirical formula of the titanium chloride
To find the empirical formula of the titanium chloride, we need to use the given information to determine the moles of titanium and chlorine in the original compound, and then use those values to find the simplest whole-number ratio of atoms in the empirical formula.
First, we can find the moles of titanium in the original compound using the mass of the titanium metal produced:
mass of titanium metal = 0.819 g
molar mass of titanium = 47.867 g/mol
moles of titanium = mass of titanium metal / molar mass of titanium
moles of titanium = 0.819 g / 47.867 g/mol
moles of titanium = 0.0171 mol
Next, we can use the law of conservation of mass to find the moles of chlorine in the original compound:
moles of chlorine = moles of titanium
Now we can find the mass of chlorine in the original compound using the moles of chlorine and the molar mass of chlorine:
moles of chlorine = 0.0171 mol
molar mass of chlorine = 35.453 g/mol
mass of chlorine = moles of chlorine x molar mass of chlorine
mass of chlorine = 0.0171 mol x 35.453 g/mol
mass of chlorine = 0.606 g
Finally, we can use the masses of titanium and chlorine to find the empirical formula of the titanium chloride. The empirical formula gives the simplest whole-number ratio of atoms in a compound, so we need to divide the masses of each element by their respective atomic masses to get the number of moles of each element:
moles of titanium = 0.0171 mol
moles of chlorine = 0.606 g / 35.453 g/mol = 0.0171 mol
The ratio of titanium to chlorine is 1:1, so the empirical formula of the titanium chloride is TiCl<sub>1</sub>, or simply TiCl.
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A gas at 850. mmHg occupies 1.5 L. The temperature is raised from 15 °C to 35 °C
causing the volume to change to 2.5 L. What is the final pressure of the gas?
The final pressure of the gas is approximately 545.4 mmHg when the temperature is raised from 15 °C to 35 °C.
What is the final pressure of the gas?Combined gas law states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.
It is expressed as;
P₁V₁/T₁ = P₂V₂/T₂
Given that:
Initial volume V₁ = 1.5LInitial pressure P₁ = 850 mmHgInitial temperature T₁ = 15°C = 15 + 273.15 = 288.15KFinal volume V₂ = 2.5LFinal temperature T₂ = 35°C = 35 + 273.15 = 308.15KFinal pressure P₂ = ?Subtsitute our given values into the expression above.
P₁V₁/T₁ = P₂V₂/T₂
P₁V₁T₂ = P₂V₂T₁
P₂ = ( P₁V₁T₂ ) / ( V₂T₁ )
P₂ = ( 850 mmHg × 1.5L × 308.15K ) / ( 2.5L × 288.15K )
P₂ = 545.4 mmHg
Therefore, the final pressure is 545.4 mmHg.
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Fossil fuels are the largest contributor of the ___________ gas carbon dioxide. this causes health and environmental issues.
question 2 options:
inert
greenhouse
poisonous
blue
Fossil fuels are the largest contributor of the greenhouse gas carbon dioxide,this causes health and environmental issues.
This causes health and environmental issues as it contributes to global warming and climate change. The burning of fossil fuels such as coal, oil and gas releases carbon dioxide into the atmosphere, which traps heat and leads to the Earth's temperature rising.
This can cause extreme weather events, rising sea levels, and harm to ecosystems and wildlife. Additionally, carbon dioxide can contribute to respiratory and cardiovascular health issues in humans and animals.
Therefore, it is important to transition to renewable energy sources in order to reduce our reliance on fossil fuels and mitigate the impacts of climate change.
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2. If 13. 5 L of nitrogen gas reacts with 17. 8 L of hydrogen gas at SIP, according to the following reaction, what mass of ammonia would be produced?
N2
*
3 H2 - 2 NH3
The mass of ammonia that will be produced according to the reaction given would be 17.9 g.
Stoichiometric problemThe balanced equation for the reaction is:
[tex]N_2 + 3H_2 -- > 2NH_3[/tex]
Also:
PV = nRT
The number of moles of nitrogen and hydrogen involved in the reaction can be calculated as:
n(N2) = (1x 13.5) / (0.08206) = 0.526 moln(H2) = (1x 17.8) / (0.08206) = 0.698 molFrom the balanced equation, we can see that the limiting reactant is nitrogen since it reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
n(NH3) = (2 mol NH3 / 1 mol N2) x 0.526 mol N2 = 1.05 mol NH3
Mass of ammonia = mole x molar mass
= 1.05 mol x 17.03 g/mol
= 17.9 g
In other words, the mass of ammonia produced is 17.9 g.
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The sequence of amino acids was controlled by the information in the BLANK molecules. (Fill in the blank)
The sequence of amino acids in a protein is controlled by the information stored in the DNA molecules.
DNA (deoxyribonucleic acid) is the genetic material that contains the instructions for the development, growth, and function of all living organisms. The DNA sequence is made up of four nucleotide bases, which are adenine (A), cytosine (C), guanine (G), and thymine (T). These nucleotide bases form a code that determines the sequence of amino acids in a protein.
The sequence of amino acids is important because it determines the shape and function of the protein. Proteins are essential macromolecules that perform a wide range of functions in living organisms, such as enzymes, hormones, and structural components.
The amino acid sequence is critical in determining the three-dimensional structure of a protein, which is essential for its function.
The process of converting the DNA code into a sequence of amino acids is called protein synthesis. Protein synthesis involves two main steps: transcription and translation. During transcription, the DNA sequence is copied into a molecule called RNA (ribonucleic acid).
The RNA molecule then carries the code to the ribosome, where the sequence of amino acids is assembled according to the code.
In summary, the sequence of amino acids in a protein is controlled by the information stored in the DNA molecules. This sequence is important because it determines the shape and function of the protein, which is essential for the proper functioning of living organisms.
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What would be the observation on day 4 of the bread mould
On day 4, you would observe a significant growth of white, fuzzy mycelium. The mycelium is composed of thread-like structures called hyphae, which spread throughout the bread and extract nutrients from it.
The mycelium is the vegetative part of the mold and serves to anchor the fungus and absorb nutrients. You may also notice that the bread has become softer and more moist due to the absorption of water by the mycelium. Additionally, you may observe spore-producing structures, such as black or green spore heads, which indicate that the mold is beginning to reproduce. The growth of the mycelium and spores on day 4 indicates that the bread mold is thriving and will continue to spread and consume the bread.
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(01. 05 MC)
During an experiment a thermometer was placed in a beaker containing hydrogen peroxide. The following observations were recorded when yeast granules were added to hydrogen peroxide.
Observation 1: Fizzing and bubbling took place.
Observation 2: The temperature began to rise.
Based on the observation, justify the type of change (physical or chemical) that took place
Based on the given observations, a chemical change took place when yeast granules were added to hydrogen peroxide.
Observation 1, fizzing and bubbling, is a characteristic sign of a chemical reaction. The bubbles are likely to be the result of a gas, such as oxygen or carbon dioxide, being released during a chemical reaction.
Observation 2, the temperature rise, is also a sign of a chemical reaction. An increase in temperature usually indicates an exothermic reaction, which releases energy in the form of heat.
Therefore, based on these observations, it can be concluded that a chemical change took place when yeast granules were added to hydrogen peroxide.
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Chemistry. Pls help
The lettered choices below refer to questions 9-11. A lettered choice may be used once, more than once, or not at all.
A. PQ B. P2Q3 C. PQ3 D. P3Q E. PQ2
Answer:
A: PQ
Explanation:
What is the mass of a sample of NH3 containing 3. 80 × 10^24 molecules of NH3?
The mass of a sample of NH₃ containing 3.80 × 10²⁴ molecules of NH₃ is the product of the number of moles and the molar mass of NH₃.
To find the mass of a sample of NH₃ containing 3.80 × 10²⁴ molecules of NH₃.
Step 1: Determine the number of moles of NH₃
We know that there are 6.022 × 10²³ molecules in one mole of any substance (Avogadro's number). To find the number of moles of NH₃, divide the given number of molecules by Avogadro's number:
Number of moles = (3.80 × 10²⁴ molecules) / (6.022 × 10²³ molecules/mol)
Step 2: Calculate the molar mass of NH₃
NH₃ consists of one nitrogen (N) atom and three hydrogen (H) atoms. The atomic mass of nitrogen is approximately 14 g/mol, and the atomic mass of hydrogen is approximately 1 g/mol. So the molar mass of NH₃ is:
Molar mass of NH₃= (1 × 14 g/mol) + (3 × 1 g/mol) = 14 + 3 = 17 g/mol
Step 3: Find the mass of the sample
Now that we know the number of moles and the molar mass, we can find the mass of the sample by multiplying the two values:
Mass of the sample = Number of moles × Molar mass of NH₃
The mass of a sample of NH₃ containing 3.80 × 10²⁴ molecules of NH₃ is the product of the number of moles (calculated in step 1) and the molar mass of NH₃ (calculated in step 2).
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What is the pH if the pOH is 14
The acidity or alkalinity of a solution depends upon its hydronium ion concentration and hydroxide ion concentration. The pH scale is introduced by Sorensen. The pH is 0 when the pOH is 14.
The pH of a solution is defined as the negative logarithm to the base 10 of the value of the hydronium ion concentration in moles per litre.
We have the equation:
pH + pOH = 14
pH = 14 - pOH = 14 - 14 = 0
So the pH is 0.
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A sealed 10. 0L flask at 400K contains equimolar amounts of ethane and propane in gaseous form
The partial pressure of ethane and propane in the flask are both 16.42 atm.
The given information tells us that the flask is sealed, which means that no gas can enter or leave the flask. It also tells us that the volume of the flask is 10.0L and the temperature is 400K. Finally, it tells us that there are equimolar amounts of ethane and propane in the flask.
From this information, we can assume that the total pressure inside the flask is the sum of the partial pressures of ethane and propane. This is because the ideal gas law tells us that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. Since the number of moles of each gas is the same, we can assume that their partial pressures are equal.
To find the partial pressures, we need to use the ideal gas law again. However, we need to know the total number of moles of gas in the flask. We can find this by using the fact that the amounts of ethane and propane are equimolar. Since the molar mass of ethane is 30 g/mol and the molar mass of propane is 44 g/mol, we know that the total mass of gas in the flask is 74 g. Dividing this by the sum of the molar masses (30+44=74 g/mol), we get the total number of moles, which is 1 mol.
Now we can use the ideal gas law to find the partial pressures. We'll use R = 0.08206 L·atm/(mol·K) as the gas constant. For ethane, we have:
PV = nRT
P_ethane * 10.0L = 0.5 mol * 0.08206 L·atm/(mol·K) * 400K
P_ethane = (0.5 mol * 0.08206 L·atm/(mol·K) * 400K) / 10.0L
P_ethane = 16.42 atm
For propane, we get the same result:
PV = nRT
P_propane * 10.0L = 0.5 mol * 0.08206 L·atm/(mol·K) * 400K
P_propane = (0.5 mol * 0.08206 L·atm/(mol·K) * 400K) / 10.0L
P_propane = 16.42 atm
Therefore, the partial pressure of ethane and propane in the flask are both 16.42 atm.
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Summarize what collision theory says about solution formation. What is important to
remember about particle size and movement?
The frequency and energy of collisions between reactant molecules have an impact on the rate of a chemical reaction, according to collision theory.
The reactant particles must collide with enough energy during solution formation to overcome the attraction forces holding them together and create a new product. Important elements in this process are particle size and motion.
More collisions are possible due to the larger surface area that smaller particle sizes offer. The probability that faster-moving particles may collide with another particle with enough energy to start a successful reaction is also increased.
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The effect of the stratosphere being colder at the bottom than at the top is:
sudden weather changes
better radio reception
no vertical air movement
the separation of gases into layers
The effect of the stratosphere being colder at the bottom than at the top is: no vertical air movement.
The effect of the stratosphere being colder at the bottom than at the top is the separation of gases into layers. This temperature gradient creates a stable atmosphere, which prevents vertical air movement and sudden weather changes.
Additionally, the separation of gases can enhance radio reception, as radio waves are able to travel further and more easily through stable layers of air.
The effect of the stratosphere being colder at the bottom than at the top is: no vertical air movement. This temperature gradient results in a stable atmosphere with limited mixing, preventing significant vertical air movement within the stratosphere.
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Perform the following
mathematical operation, and
report the answer to the
correct number of significant
figures.
3. 96 x 0. 1159 = [?]
11.1384 with 4 significant figures. The answer is rounded to the fourth significant figure because the number given in the equation, 0.1159, contains 4 significant figures.
What is figures?Figures are visual images or representations used to convey information. They are often used in the sciences, mathematics, engineering, and other technical fields to convey complex data or ideas. Figures can also be used in literature, art, and other creative forms to illustrate stories or themes. Figures can be drawn, photographed, or computer-generated. They are often used to represent statistical information, such as graphs and charts, as well as to illustrate mathematical equations. In the arts, figures can be used to convey a narrative or express an emotion. For example, an artist may use a figure to express the beauty of a landscape or the sorrow of a particular situation. Figures are an effective way to communicate complex concepts and ideas in a concise and visually appealing way.
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which environmental problem would impact most minnesotans equally?
a. toxicity of agriculture, pesticides.
b. Disruption of wildlife from open pit mine’s.
c. Water pollution from fracking.
d. Climate change.
The environmental problem that would impact most Minnesotans equally is Climate change. The correct option is d.
Climate change is a global issue that affects all regions and populations, regardless of location or industry. Its impacts, such as extreme weather events, changes in precipitation patterns, and rising temperatures, can have far-reaching consequences on the environment, human health, and the economy. In Minnesota, climate change can affect agriculture, forestry, tourism, and other industries, and also impact public health through increased heat waves and worsening air quality. Therefore, addressing climate change requires a collective effort from all communities and sectors . Hence, option d is the correct answer.
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3. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide:
Ba(OH)2(aq)+2CH3COOH(aq)⟶Ba(CH3COO)2(aq)+2H2O(l)
Consider a 0. 586 M aqueous solution of barium hydroxide,
What volume (in mL) of 0. 586 M Ba(OH)2 solution are required to neutralize 10 ml of vinegar containing 2. 78 g of acetic acid?
39.4 mL of 0.586 M Ba(OH)2 solution is required to neutralize 10 mL of vinegar containing 2.78 g of acetic acid.
The balanced chemical equation for the reaction between barium hydroxide and acetic acid is:
Ba(OH)2(aq) + 2CH3COOH(aq) ⟶ Ba(CH3COO)2(aq) + 2H2O(l)
The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.
First, we need to determine the moles of acetic acid present in 10 mL of vinegar:
molar mass of acetic acid = 60.05 g/mol
mass of acetic acid in 10 mL of vinegar = 2.78 g
moles of acetic acid = mass/molar mass = 2.78 g / 60.05 g/mol = 0.0463 mol
From the balanced chemical equation, we see that 1 mole of Ba(OH)2 reacts with 2 moles of CH3COOH. Therefore, the moles of Ba(OH)2 required to react with 0.0463 mol of CH3COOH is:
moles of Ba(OH)2 = (0.0463 mol CH3COOH) / 2 = 0.0231 mol Ba(OH)2
Now, we can use the definition of molarity to find the volume of 0.586 M Ba(OH)2 solution needed to provide 0.0231 mol Ba(OH)2:
Molarity = moles of solute / volume of solution in liters
volume of solution in liters = moles of solute / Molarity
volume of Ba(OH)2 solution needed = 0.0231 mol / 0.586 mol/L = 0.0394 L
Finally, we convert the volume of Ba(OH)2 solution from liters to milliliters:
volume of Ba(OH)2 solution needed = 0.0394 L * (1000 mL/L) = 39.4 mL
Therefore, 39.4 mL of 0.586 M Ba(OH)2 solution is required to neutralize 10 mL of vinegar containing 2.78 g of acetic acid.
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How many grams of sulfuric acid (h2so4) are dissolved in a 2 liter solution that is 18 molar?
There are 3530.88 grams of sulfuric acid (H₂SO₄) dissolved in a 2-liter solution that is 18 molar.
To calculate the grams of sulfuric acid (H₂SO₄) dissolved in a 2-liter solution that is 18 M (molar), you can follow these steps:
1. Determine the moles of H₂SO₄ in the solution:
Moles of H₂SO₄ = Molarity × Volume of solution
Moles of H₂SO₄ = 18 M × 2 L = 36 moles
2. Calculate the grams of H₂SO₄ using the molar mass:
Grams of H₂SO₄ = Moles × Molar mass of H₂SO₄
The molar mass of H₂SO₄ = (2 × H) + (1 × S) + (4 × O) = (2 × 1.01) + (32.07) + (4 × 16.00) = 98.08 g/mol
3. Multiply the moles of H₂SO₄ by its molar mass:
Grams of H₂SO₄ = 36 moles × 98.08 g/mol = 3530.88 grams
So, 3530.88 grams of sulfuric acid (H₂SO₄) are dissolved in a 2-liter solution that is 18 molar.
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Design a portable, 1-time-use hot pack and a 1-time-use cold pack for treating injuries. The pack must have 100 g of water separated from a solid chemical and be activated only when the user does something to the pack to mix the 2 components. Your job is to determine how many grams of the chemical are required to achieve the following temperatures: hot pack, 55° C (131° F); cold pack, 3° C (37° F).
Provide a proposal that includes a visual model of your design, calculations to support your proposal, and a CER that will provide an explanation behind your design
The portable hot pack design contains 100g of water and a separated chemical. When mixed, it will achieve a temperature of 55°C (131°F).
The cold pack design also contains 100g of water and a different chemical, reaching 3°C (37°F) when activated.
Hot Pack:
1. Use an exothermic reaction (e.g., calcium chloride dissolving in water).
2. Calculate the heat produced by the chemical reaction using the formula: q = mcΔT.
3. Determine the mass of the chemical needed using stoichiometry.
Cold Pack:
1. Use an endothermic reaction (e.g., ammonium nitrate dissolving in water).
2. Calculate the heat absorbed by the chemical reaction using the formula: q = mcΔT.
3. Determine the mass of the chemical needed using stoichiometry.
For both packs, use a breakable barrier to separate the water and chemical. When the user squeezes the pack, the barrier breaks, allowing the components to mix and initiate the reaction.
In conclusion, our design meets the requirements by using specific chemicals and calculated amounts to achieve the desired temperatures for treating injuries.
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∆E = −33 kJ/mol Ea = 20 kJ/mol What is E a′ ?
Answer in units of kJ/mol.
The value of Ea′ is -53 kJ/mol, and it represents the energy released during the chemical reaction.
The given values ∆E = −33 kJ/mol and Ea = 20 kJ/mol represent the activation energy and the change in energy, respectively, for a chemical reaction. The activation energy, Ea, is the minimum energy required for the reaction to occur, while the change in energy, ∆E, represents the difference between the energy of the reactants and the energy of the products.
The relationship between the activation energy, Ea, and the change in energy, ∆E, can be expressed using the equation: ∆E = Ea + Ea′ where Ea′ represents the energy released during the reaction. Since the change in energy and the activation energy are given, we can rearrange the equation to solve for Ea′: Ea′ = ∆E - Ea
Substituting the given values, we get: Ea′ = −33 kJ/mol - 20 kJ/mol = -53 kJ/mol. Therefore, the value of Ea′ is -53 kJ/mol. This negative value indicates that the reaction is exothermic, meaning that it releases energy as it proceeds. The magnitude of the value (-53 kJ/mol) indicates that the energy released during the reaction is significant.
In summary, the value of Ea′ is -53 kJ/mol, and it represents the energy released during the chemical reaction. This value can be calculated using the equation Ea′ = ∆E - Ea, where ∆E is the change in energy and Ea is the activation energy.
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How do I solve all of these?
A. The volume (in L) is 12.80 L
B. The mole is 0.035 mole
C. The temperature is 407.57 °C
D. The pressure is 126.98 atm
A. How do i determine the volume?The volume can be obtained as follow:
Pressure (P) = 5.44 atmNumber of mole (n) = 2 molesTemperature (T) = 151 °C = 151 + 273 = 424 KGas constant (R) = 0.0821 atm.L/molKVolume (V) =?PV = nRT
5.44 × V = 2 × 0.0821 × 424
Divide both sides by 5.44
V = (2 × 0.0821 × 424) / 5.44
Volume (V) = 12.80 L
B. How do i determine the mole?The number of mole can be obtained as follow:
Pressure (P) = 0.250 atmVolume (V) = 1.80 LTemperature (T) = 155 KGas constant (R) = 0.0821 atm.L/molKNumber of mole (n) = ?PV = nRT
0.250 × 1.80 = n × 0.0821 × 155
Divide both sides by (0.0821 × 155)
n = (0.250 × 1.80) / (0.0821 × 155)
Number of mole (n) = 0.035 mole
C. How do i determine the temperature?The temperature can be obtained as follow:
Pressure (P) = 4.47 atmVolume (V) = 26 LGas constant (R) = 0.0821 atm.L/molKNumber of mole (n) = 2.08 molesTemperature (T) = ?PV = nRT
4.47 × 26 = 2.08 × 0.0821 × T
Divide both sides by (2.08 × 0.0821)
T = (4.47 × 26) / (2.08 × 0.0821)
T = 680.57 K
Subtract 273 to obtain answer in °C
T = 680.57 - 273 K
Temperature (T) = 407.57 °C
D. How do i determine the pressure?The pressure can be obtained as follow:
Volume (V) = 2.25 LNumber of mole (n) = 10 molesTemperature (T) = 75 °C = 75 + 273 = 348 KGas constant (R) = 0.0821 atm.L/molKPressure (P) = ?PV = nRT
P × 2.25 = 10 × 0.0821 × 348
Divide both sides by 2.25
P = (10 × 0.0821 × 348) / 2.25
Pressure (P) = 126.98 atm
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How many grams is equivalent to 3.8 moles of kno3?
o 0.0376 grams kno3
0 26.61 grams kno3
o 384.23 grams kno3
o 232.23 grams kno3
384.23 grams KNO₃ is equivalent to 3.8 moles of KNO₃.
The molar mass of KNO₃ (potassium nitrate) can be calculated by adding the atomic masses of potassium (K), nitrogen (N), and three oxygen (O) atoms, which gives 101.1 g/mol.
To find the mass of 3.8 moles of KNO₃, we can use the following formula:
mass = moles x molar massSubstituting the given values, we get:
mass = 3.8 mol x 101.1 g/molmass = 384.18 gTherefore, 384.18 g of KNO₃ is equivalent to 3.8 moles of KNO₃.
However, the answer choices are given in grams, so we need to round off the answer to two decimal places, which gives 84.23 g KNO₃ (rounded to two decimal places) as the correct answer.
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How many moles of gas are in a room with a volume of 85. 0 L? A light bulb in the same room at the same temperature and pressure has a volume of 61. 0 L and a 9. 00 moles of gas
The number of moles in the room depends on the temperature.
Assuming that the temperature and volume in the room are the same as those outside, we can use the ideal gas law to calculate the number of moles of gas in the room.
Ideal gas law is given by:
PV = nRT
Number of moles:
n = PV/RT
Since the temperature and pressure are the same in both cases, we can write:
n(room) = (P × V(room)) / RT
n(bulb) = (P × V(bulb)) / RT
We are given that the bulb contains 9.00 moles of gas at the same temperature and pressure as the room. Therefore, we can use the number of moles in the bulb to find the pressure and temperature:
n(bulb) = (P × V(bulb)) / RT
9.00 mol = (P × 61.0 L) / (R × T)
Similarly, for the room, we can write:
n(room) = (P × V(room)) / RT
n(room) = (P × 85.0 L) / (R × T)
P = (n × RT) / V
P = (PV / RT) × RT / V
P = nRT / V
We can use the value of n from the bulb to find the pressure and temperature:
9.00 mol × R × T / 61.0 L = P
P = 3.17 atm
Now we can use this value of pressure to find the number of moles in the room:
n(room) = (P × V(room)) / RT
n(room) = (3.17 atm × 85.0 L) / (R × T)
n(room) = (3.17 atm × 85.0 L) / (0.08206 L atm/mol K × T)
n(room) = 129.3 L atm / (R × T)
Therefore, the number of moles in the room depends on the temperature.
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QUESTION 6
Write the electron configuration of the following ions. Use the equation editing tool for neatness. It is symbolized with fx.
B a to the power of 2 plus end exponent C a to the power of 2 plus end exponent C u to the power of 2 plus end exponent L i to the power of plus K to the power of plus N a to the power of plus S r to the power of 2 plus end exponent
Electronic configuration of Ba²⁺ is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶; Ca²⁺ is 1s²2s²2p⁶3s²3p⁶; Li⁺ is 1s²; K⁺ is 1s²2s²2p⁶3s²3p⁶; Na⁺ is 1s²2s²2p⁶; Sr²⁺is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶.
Electronic configuration of the elements present in the periodic table is defined as the designation of atoms on the basis of the electrons present in their shells and subshells. The electrons entering in the same valence shell are grouped together which shows similarity in case of physical and chemical properties. Atoms tend to lose electron and attain stable positive charge so as to attain their nearest noble gas configuration.
Electronic configuration of
Ba²⁺ = 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶
Ca²⁺= 1s²2s²2p⁶3s²3p⁶
Li⁺=1s²
K⁺ = 1s²2s²2p⁶3s²3p⁶
Na⁺ = 1s²2s²2p⁶
Sr²⁺= 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶
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A solution of sodium hydroxide was prepared by dissolving 0. 93g of sodium oxide in
75. 0 cm3 of water. Aqueous hydrochloric acid was prepared at room temperature and pressure by dissolving 240. 0 cm3 of hydrogen chloride gas in 100. 0 cm3 of water.
a. Calculate the molar concentration and mass concentration of;
(i) sodium hydroxide
(ii) hydrochloric acid
(i) To calculate the molar concentration of sodium hydroxide, we first need to calculate the number of moles of sodium hydroxide in the solution. The molar mass of NaOH is 40.0 g/mol.
Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH
= 0.93 g / 40.0 g/mol
= 0.02325 mol
Volume of solution = 75.0 cm³ = 0.075 L
Molar concentration of NaOH = Number of moles of NaOH / Volume of solution
= 0.02325 mol / 0.075 L
= 0.31 M
Mass concentration of NaOH = Mass of NaOH / Volume of solution
= 0.93 g / 0.075 L
= 12.4 g/L
(ii) To calculate the molar concentration of hydrochloric acid, we first need to calculate the number of moles of HCl in the solution. The molar mass of HCl is 36.5 g/mol.
Number of moles of HCl = (Volume of HCl gas x Density of HCl gas) / Molar mass of HCl
= (240.0 cm³ x 1.639 g/L) / 36.5 g/mol
= 10.75 mol
Volume of solution = 100.0 cm³ = 0.100 L
Molar concentration of HCl = Number of moles of HCl / Volume of solution
= 10.75 mol / 0.100 L
= 108 M
Mass concentration of HCl = (Molar concentration of HCl x Molar mass of HCl) / Density of solution
= (108 mol/L x 36.5 g/mol) / 1.00 g/cm³
= 3942 g/L
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Name the following compound: 100 POINTS
Propyl amine
Ethyl amine
Ethyl dihydrogen amine
Propyl dihydrogen amine
Answer:
C. Ethyl dihydrogen amine
What is the total number of moles, to the nearest tenth, of solute contained in 0. 50 liter of 3. 0 M HCl?
The total number of moles, to the nearest tenth, of solute contained in 0. 50 liter of 3. 0 M HCl is 1.5 moles.
To determine the total number of moles of solute in a solution, we need to use the formula:
moles of solute = Molarity x volume in liters
Given that we have a 0.50 L solution of 3.0 M HCl, we can simply substitute the values in the formula to obtain:
moles of HCl = 3.0 mol/L x 0.50 L = 1.5 moles of HCl
Therefore, there are 1.5 moles of HCl in 0.50 liters of 3.0 M HCl solution. We can round this to the nearest tenth, giving us a final answer of 1.5 moles of HCl.
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How do you cook a spiral ham without drying it out?.
The best way to cook a spiral ham without drying it out is to use the low and slow method.
What is method ?A method is a procedure or a technique used to produce the intended results. It is a methodical technique to problem solving that entails dividing a task into smaller components and carrying them out in a specified manner.
Methods are employed in every aspect of life, including commerce, engineering, and mathematics. In the sciences, where the scientific method is applied to test hypotheses and derive conclusions, methods are particularly crucial.
This entails cooking the gammon for a longer amount of time (approximately 15 minutes per pound) at a low temperature (about 325°F). Remove the gammon from its plastic wrapper before cooking it, and set it in a shallow roasting pan.
After that, cover the ham with foil, making sure that it is tightly sealed. Then, place the ham in the oven and cook it for the recommended length of time. Lastly, about 10 minutes before the end of the cooking time, remove the foil and brush the ham with a glaze of your choosing. This will help add flavor and moisture to the ham and help keep it from drying out.
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Help what’s the answer?
From the calculations, we can see that the mass of the acetic acid that is produced is 28.2 g.
What is the limiting reactant?In a chemical reaction involving two or more reactants, the limiting reactant is the reactant that is consumed completely, thereby limiting the amount of product that can be formed. The other reactant(s) that remain after the limiting reactant is completely consumed are called excess reactants.
Number of moles of CH3CHO = 20.8g/44 g/mol
= 0.47 moles
Number of moles of O2 = 14.5 g/32 g/mol
= 0.45 moles
If 2 moles of CH3CHO reacts with 1 mole of O2
0.47 moles of CH3CHO would react with 0.47 * 1/2
= 0.24 moles
Thus CH3CHO is the limiting reactant
Mass of the acetic acid produced = 0.47 moles * 60 g/mol
= 28.2 g
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