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Roman55 [17]
3 years ago
12

Select all that apply. At STP conditions, 0.50 moles of CO(g), Cl2(g), O2(g): will contain the same number of molecules, will co

ntain the same number of atoms, will occupy the same volume, will have the same g.f.w.
Chemistry
1 answer:
Mariana [72]3 years ago
6 0
1 mol is (always) the same number of units: 6.02 * 10^23 units.


So 0.5 moles  of any gas has the same number of molecules.


Also, we know by the ideal gas laws that a give number of molecules of any gas will occupy the same volume.


Given that the three gases have the same number of atoms in the molecular fomula (2), 0.5 moles will also have the same number of atoms.


g.f.w. stands for grams formula weight and that is different for all of them, because gfw is calcualted from the atomics masses of each atom in the molecule.


Then at STP (standard temperature and pressure) conditions 0.50 moles of any of the gas:


- will contain the same number of molecules,

- will contain the same number of atoms

- will occupy the same volume.

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A sample of a gas has a volume of 852 mL at 298 K. If the gas is cooled to 200K, what would the new volume be?
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Assuming constant pressure, we can solve this problem by using <em>Charles' law</em>, which states that at constant pressure:

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Where in this case:

  • V₁ = 852 mL
  • T₂ = 200 K
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  • T₁ = 298 K

We <u>input the data</u>:

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And <u>solve for V₂</u>:

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4 0
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If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
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<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

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The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0
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