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KatRina [158]
3 years ago
15

Select all statements that are correct:____

Chemistry
1 answer:
topjm [15]3 years ago
4 0

Answer:

A, C and D are correct.

Explanation:

Hello.

In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:

P_{vap}^{solution}=x_{solvent}P_{solvent}

Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.

Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.

Best regards.

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Yes, if it’s a parallel circuit the wires are two different wires so it will light because that bulb isn’t connected to the one that went out
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Without using Appendix B, predict the sign of ΔS° for(b) KBr(s)→ KBr(aq)
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Since the sign is positive, the entropy increased by 88.48 J/K.

Examine the phases of the species present to determine whether a physical or chemical process will cause an increase or decrease in entropy. Keep in mind "Silly Little Goats" to aid you in telling.

[1 Sf K+1 + 1 Sf Br-1 (aq)] ([1Sf(KBr (s))])

[1(102.5) + 1(82.42)] - [1(96.44)] = 88.48 J/K

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A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t
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Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)

So, moles of product formed are calculated as follows.

\frac{3}{2} \times 0.17 mol \\= 0.255 mol

Hence, the given data is as follows.

n_{1} = 0.17 mol,      n_{2} = 0.255 mol

V_{1} = 19.5 L,         V_{2} = ?

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

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