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Bezzdna [24]
3 years ago
7

A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L. If the containe

r is evacuated (all of the gas removed), sealed, and then allowed to warm to room temperature T = 298 K so that all of the solid CO2 is converted to a gas, what is the pressure inside the container?
Chemistry
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

The pressure inside the container is 6.7 atm

Explanation:

We have the ideal gas equation: P x V = n x R x T

whereas, P (pressure, atm), V (volume, L), n (mole, mol), R (ideal gas constant, 0.082), T (temperature, Kelvin)

Since the container is evacuated and then sealed, the volume of the body of gas is the volume of the container.

So we can calculate the pressure by

P = n x R x T / V

where as,

n = 41.1 g / 44 g/mol = 0.934 mol

Hence P = 0.934 x 0.082 x 298 / 3.4 L = 6.7 atm

You might be interested in
The solubility of two slightly soluble salts of M^2+ , MA and MZ_2, are the same, 4 x 10^â4 mol/L.
LekaFEV [45]

Answer:

a) MZ₂

b) They have the same concentration

c) 4x10⁻⁴ mol/L

Explanation:

a) The solubility (S) is the concentration of the salt that will be dissociated and form the ions in the solution, the solubility product constant (Kps) is the multiplication of the concentration of the ions elevated at their coefficients. The concentration of the ions depends on the stoichiometry and will be equivalent to S.

The salts solubilization reactions and their Kps values are:

MA(s) ⇄ M⁺²(aq) + A⁻²(aq) Kps = S*S = S²

MZ₂(aq) ⇄ M⁺²(aq) + 2Z⁻(aq) Kps = S*S² = S³

Thus, the Kps of MZ₂ has a larger value.

b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.

c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:

8x10⁻⁴ mol/2 L = 4x10⁻⁴ mol/L.

7 0
3 years ago
All gaseous mixtures are solutions.
Gre4nikov [31]
The statement "<span>All gaseous mixtures are solutions." is true. This is because there are a number of gaseous molecules present in a volume of gas and they are considered solutions.</span>
6 0
3 years ago
The Law of Superposition is most relevant when studying which type of rock?
bagirrra123 [75]

Answer:

igneous or metamorphic

Explanation:

Those two are sorta relvant

8 0
3 years ago
Read 2 more answers
Use the half reaction method to balance the following reaction in acid:
sergij07 [2.7K]

Answer:

I2 + 2S2O3^2- --------> S4O6^2- + 2I^-

Explanation:

The correct equation is;

2Na2S2O3 + I2 → Na2S4O6 + 2NaI

Oxidation half equation;

2S2O3^2- --------> S4O6^2- + 2e-

Reduction half equation;

I2 + 2e ----------> 2I^-

Overall balanced redox reaction equation;

I2 + 2S2O3^2- --------> S4O6^2- + 2I^-

6 0
3 years ago
Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O
Feliz [49]

Answer:

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O

Explanation:

PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O

It will be easiest to balance this equation by the ion-electron method.

1. Write the ionic equation

PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O

2. Eliminate H⁺, H₂O, and spectator ions

PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺  

3. Separate the skeleton equation into two half-reactions.

PbO₂  ⟶ Pb²⁺  

Mn²⁺ ⟶ MnO₄⁻

4. Balance all atoms other than H and O

Done

5. Balance O by adding water molecules to the deficient side

           PbO₂  ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻

6. Balance H by adding H⁺ ions to the deficient side.

  PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺

7. Balance charge by adding electrons to the deficient side.

PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O

     Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-

 

8. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]

     2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]

9. Add the two half-reactions.

                          5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O

<u>                                    2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻                    </u>

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>

becomes

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

11. Add the missing spectator ions

5PbO₂ + 2Mn²⁺    + 4H⁺                              ⟶            5Pb²⁺   + 2MnO₄⁻ + 2H₂O

            + 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻       +2SO₄²⁻ + 6NO₃⁻ + 2H⁺

becomes

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

12. Check that all atoms are balanced.

\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}

Everything checks. The balanced equation is

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

7 0
4 years ago
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