Answer:
3.77 mg of K-40 decayed into Ar-40.
Data:
1) K-40, Ca-40, Ar-40: all three have the same atomic mass
2) 90% of the potassium-40 will decay into calcium-40
3) 10% of the potassium-40 will decay into argon-40.
4) K-40 inside the rock = 0.81 mg
5) Ar-40 trapped = 0.377 mg
Soltuion:
1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed
=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg
=> x = 0.377 mg / 0.1 = 3.77 mg
That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.
Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.
2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>
mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg
3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg
4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg
5) amount of K-40 that decayed into Ar-40 as percent
% = [3.77 mg / 4.58mg] * 100 = 82.31 %.
Why does reducing solute particle size increase the speed at which the solute
dissolves in water?
A. It makes the temperature of the water significantly higher.
B. It exposes morg of the solute to the water molecules.
C. It makes the water molecules move around faster.
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D. It raises the pressure of the water molecules on the solute.
Answer: B. It exposes more of the solute to the water molecules.
Explanation:
Rate of a reaction is dependent on following factors:
a) adding a catalyst: Adding a catalyst increases the rate of reaction
b) reducing the surface area: The rate of the reaction will decrease by reducing the surface area
c) raising the temperature: Increasing the temperature increases the rate of the reaction.
d) increasing the amount of reactants : Increasing the amount of reactants increases the rate of reaction.
If the reactants are present in smaller size, more reactants can react as decreasing the size increases the surface area of the reactants which will enhance the contact of molecules. Hence, more products will form leading to increased rate of reaction.
Thus reducing solute particle size increase the speed at which the solute dissolves in water by exposing more of the solute to the water molecules.
Answer:
B. It exposes more of the solute to the water molecules.
How many moles of an ideal gas are in a tank with a volume of 22.9 L at pressure of 14.297 atm at 12°C? ( Show work and units )
Answer:
14 moles
Explanation:
For an Ideal gas,
PV = nRT...................... Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = Molar gas constant.
make n the subject of the equation
n = PV/RT.................. Equation 2
Given: P = 14.297 atm, V = 22.9 L = 22.9 dm³, T = 12 °C = (12+273) K = 285 K.
Constant: R = 0.082 atm.dm³/K.mol
Substitute these values into equation 2
n = (14.297×22.9)/(285×0.082)
n = 327.4013/23.37
n = 14.009
n ≈ 14 moles
To lift fingerprints from a crime scene, a solution of silver nitrate is sprayed on a surface to react with the sodium chloride left behind by perspiration. What is the molarity of a silver nitrate solution if 42.8 mL of it reacts with excess sodium chloride to produce 0.148g of precipitate according to the following reaction?
AgNO3(aq)+NaCl(aq) --> AgCl(s)+NaNO3(aq)
A) 2.41 x 10^-2 M
B) 0.0229 M
C) 6.66 x 10^-2 M
D) 3.2 x 10^-3 M
E) 2.29 x 10^2 M
Answer:
A) 2.41 * 10^-2 M
Explanation:
Here we go from grams of percipitate to moles of Silver nitrate because we already know NaCl is excess so we don't care about that.
[tex]0.148gAgCl * \frac{1mol AgCl}{143.32 g AgCl} * \frac{1 mol AgNO3}{1 mol AgCl}\\ = 0.00103 mol AgNO3[/tex]
Now that we have moles we know molarity is moles/litres so we just plug it in:
Molarity = mol/litres = 0.00103/0.0428 L = 0.0241 M or 2.41 * 10^-2 M
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: →+2IO−3aq+12H+aq5Cos+I2s+6H2Ol5Co+2aq Suppose the cell is prepared with 6.64 M IO−3 and 1.54 M H+ in one half-cell and 7.82 M Co+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Answer:
E = 1.47
Explanation:
To do this, you need to apply the Nerst equation which is the following:
E = E° - RT/nF lnQ (1)
Where:
E: cell voltage
E°: Standard potential reduction
R: universal constant
T: temperature of the system
n: number of electrons transfered during the reaction
F: Faraday constant.
Q: Equilibrium constant
However, as the reaction is taking place at 25 °C, and R and F have constant values, we can reduce the above expression to the following:
E = E° - 0.05916/n lnQ (2)
We can get the value of Q because it has to do with the reaction which is the following:
2IO₃⁻(aq) + 12H⁺(aq) + 5Co(s) ----------> I₂(s) + 5Co²⁺(aq) + 6H₂O(l)
Now, using only the aqueous state the expression of Q will be:
Q = [Co²⁺]⁵ / [H⁺]¹² [IO₃⁻]²
Replacing the values we have:
Q = (7.82)⁵ / (1.54)¹² * (6.64)²
Q = 3.728
Knowing this, all we need to know now is the standard potential reduction of the reaction. To do so, we need to write the two semi equations of reduction and oxidation:
2IO₃⁻ + 12H⁺ + 10e⁻ ---------> I₂ + 6H₂O E₁° = 1.20 V
5Co ---------> 5Co²⁺ + 10e⁻ E₂° = 0.28 V
E° = 1.2 + 0.28 = 1.48 V
Now that we have all the values (n = 10) we can write now the nernst equation to calculate the cell voltage:
E = 1.48 - 0.05916/10 ln (3.728)
E = 1.48 - 0.005916 (1.315872)
E = 1.47 V
This will be the cell voltage
It is usually assumed that an action potential begins immediately at the cathode. If this were true, both methods for calculating conduction velocity would provide the same answer. However, when a strong stimulus intensity is used, the action potential may begin some distance away from the cathode. Under these conditions, the difference method would be more accurate.Did you observe any important difference between the conduction velocity values?
Answer: Provided in the explanation section
Explanation:
Our questions says that:
It is usually assumed that an action potential begins immediately at the cathode. If this were true, both methods for calculating conduction velocity would provide the same answer. However, when a strong stimulus intensity is used, the action potential may begin some distance away from the cathode. Under these conditions, the difference method would be more accurate.Did you observe any important difference between the conduction velocity values?
Answer to this :
By using the difference method, you subtract out any "uncertainties" involved in the measurement of latencies. Say for example, saw we are uncertain as to where the AP's are actually originating within the vicinity of the stimulating electrodes, this "error" will be introduced into both latency measurements, and therefore subtracted out when performing a difference method calculation.
However, the difference method is only experimentally sound when one is dealing with the same population of nerve fibres over the recording electrodes used, which is not the case with the sciatic nerve, as it is a short nerve, and thin at one end.
The non-uniformity of the nerve, and the difficulty in making accurate measurements of very small distances and latencies are principal points to consider when making conduction velocity measurements. Naturally if the nerve studied were longer and more uniform, we would improve the accuracy of our calculations.
cheers i hope this helped !!!!
A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric acid will: (Assume that the volume does not change upon the addition of hydrochloric acid.)
a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity
Answer:
Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.
Explanation:
The pH of the buffer solution before the addition of HCl is:
[tex]pH = pKa + log(\frac{[KF]}{[HF]})[/tex]
[tex]pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04[/tex]
The hydrochloric acid added will react with the potassium fluoride as follows:
H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)
The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:
[tex] \eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol [/tex]
[tex] \eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles [/tex]
The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:
[tex] \eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles [/tex]
Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.
We can calculate the pH after the addition of HCl:
HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)
The number of moles of HF after the reaction of KF with HCl is:
[tex] \eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles [/tex]
And the concentration of HF after the reaction of KF with HCl is is:
[tex] C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L [/tex]
Now, from the equilibrium of equation (1) we have:
[tex] Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]} [/tex]
[tex] Ka = \frac{x^{2}}{0.531 - x} [/tex] (2)
By solving equation (2) for x we have:
x = 0.0187
Finally, the pH after the addition of HCl is:
[tex] pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73 [/tex]
Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.
I hope it helps you!
What is kinetic energy?
a. The energy of change
b. The energy of distance or volume
c. The energy of motion
d. The energy of position or composition
Answer:
C. The energy of motion
Explanation:
Kinetic energy is the energy associated with the movement of objects.
The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role.
Example of Kinetic Energy:
1. An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.
// have a great day //
Kinetic energy is the energy of motion.
So kinetic energy is present when objects move.
For example, a snowball rolling down a mountain.
How much heat is liberated (in kJ) from 249 g of silver when it cools from 87 °C to 26 °C? The heat capacity of silver is 0.235 Jg^{-1} °C^{-1}. Note, "heat liberated" implies that the change in heat is negative. Enter a positive number.
Answer:
q = - 3.569KJ 0r 3.569KJ Liberated heat (signifying the change in heat is negative)
Explanation:
liberated heat implies that change in heat is negative , therefore
q = -m c ΔT
where, m = mass of the Silver = 249 g
c = specific heat capacity of Silver = 0.235 Jg^{-1} °C^{-1
ΔT = change in temperature = 87°C- 26 °C= 61°C
q = -m x c x ΔT
= - 249 x 0.235 x 61 = - 3569.415J rounded to -3569J
Changing to KJ becomes= -3569/1000= - 3.569 KJ
q = - 3.569KJ 0r 3.569KJ liberated heat.
Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUME of the equilibrium system is suddenly decreased at constant temperature: The value of Kc A. increases. B. decreases. C. remains the same. The value of Qc A. is greater than Kc. B. is equal to Kc. C. is less than Kc. The reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. The number of moles of Cl2 will: A. increase. B. decrease. C. remain the same.
Answer:
The value of Kc C. remains the same.
The value of Qc C. is less than Kc.
The reaction must: A. run in the forward direction to reestablish equilibrium
The number of moles of Cl2 will B. decrease.
Explanation:
Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.
A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.
But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.
As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:
PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)
The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.
By decreasing the volume, and so that Kc remains constant, being:
[tex]Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume[/tex]
where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂
so, the number of moles of Cl₂ should decrease.The number of moles of Cl2 will B. decrease.
If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, the value of Qc C. is less than Kc.
A box sits on a table. A short arrow labeled F subscript N = 100 N points up. A short arrow labeled F subscript g = 100 N points down. A long arrow labeled F subscript P = 75 N points right. A short arrow labeled F subscript f = 10 N points left. What is the net force acting on the box? 285 N 185 N 85 N 65 N
Answer:65N
Explanation:75-10=65
100-100=0
Therefore=65n
Answer:
the answer is 65N
Write the condensed structural formula of the ester formed when each Of the following reacts with methanol. For example, the ester formed when propanoic acid (CH3CH2COOH) reacts With methanol (HOCH3) is CH3CH2COOCH3.
a. acetic acid (CH3COOH)
b. butanoic acid
Answer:
1. Methyl ethanoate i.e CH3COOCH3
2. Methyl butanoate i.e CH3CH2CH2COOCH3
Explanation:
The reaction between an organic acid and alcohol is called esterification in which an ester is formed along side with water.
Thus, the name ester formed can be obtained as follow:
1. Reaction of acetic acid, CH3COOH with methanol, CH3OH.
CH3COOH + HOCH3 –> CH3COOCH3 + H2O
The name of the ester formed is methyl ethanoate i.e CH3COOCH3
2. Reaction of butanoic acid, CH3CH2CH2COOH with methanol, CH3OH.
CH3CH2CH2COOH + HOCH3 —> CH3CH2CH2COOCH3 + H2O
The name of the ester formed is methyl butanoate i.e CH3CH2CH2COOCH3
What is the reactant(s) in the chemical equation below?
300(g) + Fe2O3(s) → 2Fe(s) + 3C02(9)
O A. 2Fe(s)
B. Fe2O3(s)
C. 2Fe(s) + 3C02(9)
D. 300(g) + Fe2O3(s)
SUBMIT
Answer:
I think it's D. 300(g) + Fe2O3(s)
Consider the titration of 1.0 L of 1.0 M NH3 with 1.0 M HCl. Which of the following correctly describe(s) how the pH would be calculated at each of the following additions of HCl? I: At 0 L HCl, pH would be calculated based on the concentration and Kb of the weak base II: At 1 L HCl, pH would be calculated based on the concentration and Ka of the conjugate acid III: At 2 L HCl, pH would be calculated based on the concentration of excess acid in solution Group of answer choices I and III Only II Only I I, II, and III II and III
Answer:
I, II, and III
Explanation:
In the titration of NH₃ with HCl:
NH₃ + HCl → NH₄⁺ + Cl⁻
Where NH₃ is the weak base and NH₄⁺ is the conjugate acid.
I: At 0 L HCl, pH would be calculated based on the concentration and Kb of the weak base: At 0L of HCl, you will have just NH₃ in solution. That means you would calculate the pH just from the concentration of the weak base using Kb. That means I is true.
II: At 1 L HCl, pH would be calculated based on the concentration and Ka of the conjugate acid: When you add 1L of HCl, you will have in solution just NH₄⁺, the conjugate acid. That means you would calculate the pH of the solution just with the Ka of the conjugate acid and its concentration. II is true.
III: At 2 L HCl, pH would be calculated based on the concentration of excess acid in solution: At 2L of HCl solution, you have HCl in excess in the solution. As HCl is a strong acid, the pH would be affected in the big way by this concentration in excess. III is true.
what's the answer to this question
Answer:
D
Explanation:
From given choices it is D.
Why would it be important to use a homogeneous mixture instead of a heterogeneous mixture for a
specific application?
A homogeneous mixture has different properties throughout, but this would not
affect the use of a material for a specific application.
A homogeneous mixture has the same properties for every part of the mixture, so
the quality of the material is consistent.
A homogeneous mixture has the same appearance throughout, even if it has
different properties.
A homogeneous mixture has all of the required components, and this is the most
important part of using a material for a specific application.
The correct answer is B. A homogeneous mixture has the same properties for every part of the mixture, so the quality of the material is consistent.
Explanation:
In a homogenous mixture, all the components are equally integrated; this means the final result in the case of a material is a material that looks uniform and also has the same properties in every part of section. This homogeneity and consistency are especially important if you are testing the properties of the material or using it for a specific application because if you use a homogenous mixture the quality and properties will be consistent and thus you can obtain consistent results.
Select all the true statements. Group of answer choices The +3 oxidation state is characteristic of the actinides. All actinides are radioactive. Cerium (Ce) rnakes 100th in abundance (by mass %). Valence-state electronegativity is when a metal with a positive oxidation state has a greater attraction for the bonded electrons (thus a higher electronegativity) than it does when it has a 0 oxidation state. The actinides are silvery and chemically reactive. The lanthanides are in Period 7.
Answer:
The +3 oxidation state is characteristic of the actinides.
All actinides are radioactive.
Cerium (Ce) rnakes 100th in abundance (by mass %).
The actinides are silvery and chemically reactive.
In the periodic table,the The +3 oxidation state is characteristic of the actinides. All actinides are radioactive.The actinides are silvery and chemically reactive.
What is periodic table?Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.
It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.
Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.
Learn more about periodic table,here:
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According to Rutherford's nuclear theory, the core of an atom (nucleus) contains most of the_____of an atom and is______, so the majority of the mass of a fluorine atom cannot be due to its nine electrons. According to Rutherford's nuclear theory, most of the volume of an atom is empty space, so the volume of a hydrogen atom cannot be mostly due to the proton. According to Rutherford's nuclear theory, the number of negatively charged particles outside the nucleus is_____the number of positively charged particles within the nucleus, so a nitrogen atom has 7 protons and 7 electrons, while a phosphorous atom cannot have 15 protons and 150 electrons.
Answer:
matter, dense, equal
Explanation:
According to Rutherford's nuclear theory, the core of an atom (nucleus) contains most of the matter of an atom and is dense, so the majority of the mass of a fluorine atom cannot be due to its nine electrons. According to Rutherford's nuclear theory, most of the volume of an atom is empty space, so the volume of a hydrogen atom cannot be mostly due to the proton. According to Rutherford's nuclear theory, the number of negatively charged particles outside the nucleus is equal the number of positively charged particles within the nucleus, so a nitrogen atom has 7 protons and 7 electrons, while a phosphorous atom cannot have 15 protons and 150 electrons.
3. Use the balanced chemical equation from the last question to solve this situation: You combine 0.5 grams of Na2CO3 with excess CaCl2. How many grams of NaCl would you expect this reaction to produce? Show all work below. g
Answer:
0.27 g
Explanation:
The reaction equation:
[tex]Na_{2} CO_{3} + CaCl_{2}[/tex] → [tex]2NaCl + CaCO_{3}[/tex]
106g of Na2CO3 - 1 mole
0.5g of Na2CO3 = 0.5 ÷ 106
= 0.0047 moles.
1 mole of NaCl - 58.5
⇒ 0.0047 moles = 0.0047 × 58.5
= 0.27g.
When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.
We combine 0.5 grams of Na₂CO₃ with excess CaCl₂ and we want to know the mass of NaCl produced. This is a stoichiometry problem.
What is stoichiometry?Stoichiometry refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.
First, we will write the balanced chemical equation.
Na₂CO₃ + CaCl₂ ⇒ 2 NaCl + CaCO₃
We will consider the following relationships.
The molar mass of Na₂CO₃ is 105.99 g/mol.The molar ratio of Na₂CO₃ to NaCl is 1:2.The molar mass of NaCl is 58.44 g/mol.[tex]0.5 g Na_2CO_3 \times \frac{1molNa_2CO_3}{105.99gNa_2CO_3} \times \frac{2molNaCl}{1molNa_2CO_3} \times \frac{58.44gNaCl}{1molNaCl} = 0.6gNaCl[/tex]
When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.
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Which of the following is not a correct statement regarding the energy in a chemical bond? It is stored between atoms. It is known as bond energy. It is energy associated with motion. It has a fixed quantity.
Answer:it has fixed quantity
Explanation:energy between chemical bonds is hard to measure
Answer:
It has a fixed quantity.
Explanation:
A 12.00g sample of MgCl2 was dissolved in water. 0.2500mol of AgNO3 was required to precipitate all the chloride ions from the solution. Calculate the purity (as a mass percentage) of MgCl2 in the sample. Your answer should have four significant figures (round to the nearest hundredth of a percent).
Answer:
[tex]Purity=99\%[/tex]
Explanation:
Hello,
In this case, the undergoing precipitation reaction is:
[tex]MgCl_2+2AgNO_3\rightarrow Mg(NO_3)_2+2AgCl[/tex]
Thus, for the 0.2500 moles of silver nitrate, the following mass of magnesium chloride is consumed (consider their 2:1 molar ratio):
[tex]m_{MgCl_2}=0.2500molAgNO_3*\frac{1molMgCl_2}{2molAgNO_3} *\frac{95.2gMgCl_2}{1molMgCl_2} \\\\m_{MgCl_2}=11.90gMgCl_2[/tex]
Therefore, the purity of the sample is:
[tex]Purity=\frac{11.90g}{12.00g}*100\%\\ \\Purity=99\%[/tex]
Best regards.
Answer: 99. 17%
Explanation:
MgCl2(aq)+2AgNO3(aq)⟶2AgCl(s)+Mg(NO3)2(aq)
(0.2500 mol AgNO3 × 1 mol (MgCl2) /2 mol (AgNO3) × 95.211 g MgCl2 /1 mol MgCl2)
divided by 12.00 g sample = 0.99178 X 100 ≈ 99.18%
What is the scientific explanation of boiling point elevation?
Explanation:
Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water.
Hope it was helpful
An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.323 M HI, 4.34E-2 M H2 and 4.34E-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.228 mol of HI(g) is added to the flask?
Answer:
[HI] = 0.5239M
[H₂] = 7.05x10⁻²
[I₂] = 7.05x10⁻²
Explanation:
The reaction of HI to produce H₂ and I₂ is:
2HI → H₂ + I₂
Where K of reaction is defined as:
K = [H₂] [I₂] / [HI]²
Replacing with the concentrations of the gases in equilibrium:
K = [4.34x10⁻²] [4.34x10⁻²] / [0.323] ²
K = 0.0181
If you add 0.228 mol = 0.228M (Because volume of the flask is 1.0L), the concentration when the system reaches the equilibrium are:
[HI] = 0.228M + 0.323M - X = 0.551M - X
[H₂] = 4.34x10⁻² + X
[I₂] = 4.34x10⁻² + X
Where X is reaction coordinate.
Replacing in K formula:
K = 0.0181 = [4.34x10⁻²+ X] [4.34x10⁻²+ X] / [0.551 - X] ²
0.0181 = 0.00188356 + 0.0868 X + X² / 0.303601 - 1.102 X + X²
0.005495 - 0.01995 + 0.0181X² = 0.00188356 + 0.0868 X + X²
0 = -0.003611 + 0.10675X + 0.9819X²
Solving for X:
X = -0.136 → False solution, there is no negative concentrations.
X = 0.0271 → Right solution.
Replacing for concentrations of each species:
[HI] = 0.5239M[H₂] = 7.05x10⁻²[I₂] = 7.05x10⁻²Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimix blend, the partial pressures of each gas are 55.0 atm oxygen, 90.0 atm nitrogen, and 50.0 atm helium. What is the percent oxygen (by volume) in this trimex blend
Answer:
The correct answer is 28.2 %.
Explanation:
Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,
Ptotal = PO2 + PN2 + PHe
= 55 + 90 + 50
= 195 atm
The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100
Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,
%VO2 = (55/195) * 100 = 28.2%
%VN2 = (90/195) * 100 = 46%
%VHe = (50/195) * 100 = 25.64%
Hence, the percent oxygen by volume present in the blend is 28.2 %.
A 60 g piece of aluminum at 20°C is cooled to -196°C by placing it in a large container of liquid nitrogen at that temperature. How much nitrogen is vaporized? (Assume that the specific heat of aluminum is constant and is equal to 0.90 kJ/kg·K and that the vaporized nitrogen's temperature does not change.)
Answer:
0.0586 kg
Explanation:
From the question,
Heat lost by the aluminium = heat gain by nitrogen.
CM(t₁-t₂) = cm................... Equation 1
Where C = Specific heat capacity of aluminum, M = mass of aluminum, t₂ = Final temperature, t₁ = initial temperature, c = latent heat of vaporization of nitrogen, m = mass of nitrogen.
make m the subject of the equation
m = CM(t₁-t₂)/c................ Equation 2
Given: C = 900 J/kg.K, M = 60 g = 0.06 kg, t₁ = 20 °C, t₂ = -196 °C
Constant: c = 199200 J/kg
Substitute these values into equation 2
m = 900×0.06×[20-(-196)]/199200
m = 900×0.06×216/199200
m = 0.0586 kg.
Answer:
[tex]m_{N_2}=58.6gN_2[/tex]
Explanation:
Hello,
In this case, for an average temperature of -176 °C, the vaporization enthalpy of liquid nitrogen is 199.2 J/g, thus, we first compute the heat lost by the aluminium by considering it cooled mass, specific heat and temperature change:
[tex]Q=mCp(T_2-T_1)=60g*0.90\frac{J}{g\°C}*(-196-20)\°C\\ \\Q=-11664J[/tex]
Next, heat lost by the aluminium is gained by the nitrogen:
[tex]-Q_{Al}=Q_{N_2}=11664J[/tex]
Therefore, the vaporized nitrogen is:
[tex]m_{N_2}=\frac{Q_{N_2}}{\Delta H_v}=\frac{11664J}{199.2J/g}\\\\m_{N_2}=58.6gN_2[/tex]
Best regards.
How many grams of Li are there in 1.39 moles of Li?
Answer:
6.941 grams
Explanation:
The molecular formula for Lithium is Li. The Si base unit for amount of substance is mole. 1 mole is equal to 1 moles Lithium, or 6.941 grams
Answer:
9.647989999999998
Explanation:
You are required to prepare 500 ml of a 6.00 M solution of HNO3 from a stock solution of 12.0 M. Describe in detail how you would go about preparing this solution. Clearly state the volume of stock solution used, the glassware's used and the procedure.
Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Explanation:
According to the dilution law,
[tex]C_1V_1=C_2V_2[/tex]
where,
[tex]C_1[/tex] = concentration of stock solution = 12.0 M
[tex]V_1[/tex] = volume of stock solution = ?
[tex]C_2[/tex] = concentration of diluted solution= 6.00 M
[tex]V_2[/tex] = volume of diluted acid solution = 500 ml
Putting in the values we get:
[tex]12.0\times V_1=6.00\times 500[/tex]
[tex]V_1=250ml[/tex]
Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Convert 32 K to degrees Celsius.
Answer:
32 K is approx. -241.15° C
Element M reacts with oxygen to form an oxide with the formula MO. When MO is dissolved in water, the resulting solution is basic. Element M could be ________.Element M reacts with oxygen to form an oxide with the formula MO. When MO is dissolved in water, the resulting solution is basic. Element M could be ________.arsenicgermaniumchlorinecalciumselenium
Answer:
Calcium
Explanation:
Since the element reacts with oxygen to form an oxide with the formula MO, the charge on the element is +2.
Also, since the oxide MO when dissolved in water is basic, the metal is an alkali earth metal.
From the above conditions;
The metal is not arsenic because arsenic is a metalloid has the following oxides As₂O₃ and As₃O₅ and are respectively amphoteric and acidic in nature
The metal is not germanium because is a metalloid and even though germanium oxide has the formula GeO₂, it is amphoteric.
The metal is not chlorine because chlorine is a non-metal
The metal is definitely calcium because calcium oxide has the formula CaO and calcium is an alkaline earth metal.
The metal is not selenium because selenium is anon-meal and its oxide has the formula Se0₂ and is acidic
In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How many moles of CaF2 can dissolve in 100 g of water? If the density of a saturated solution of CaF2 is 1.00 g/mL, how many moles of CaF2 will dissolve in exactly 1.00 L of solution?
Answer:
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
Explanation:
First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:
Ca: 40 g/moleF: 19 g/moleSo the molar mass of CaF₂ is:
CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?
[tex]moles=\frac{0.0016 grams*1 mole}{78 grams}[/tex]
moles=2.05*10⁻⁵
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?
[tex]mass of CaF_{2}=\frac{1000 mL*1g}{1mL}[/tex]
mass of CaF₂= 1000 g
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?
[tex]moles=\frac{1000 grams*1 mole}{78 grams}[/tex]
moles=12.82
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
A chemist needs to make an acidic solution that is 0.25 M in acetic acid (HC 2 H 3 O 2 ). If she plans to prepare 250mL of solution, how much acetic acid must she use to prepare the solution.
Answer:
15g
Explanation:
Molar mass of acetic acid = 60g/mol
Molarity = mass ÷ molar mass
mass = 0.25 × 60
= 15g
∴ 15g of acetic acid weighed into 1L volumetric flask would give 0.25M of acetic acid.
Alternatively, if the solution is being prepared from a stock solution of known concentration, (say 500mL of 0.5M solution), the formular C2V1 =C2V2 can be used to find the dilution volume.