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kvasek [131]
3 years ago
13

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

? Please note this question has 1 submission for each part. (a) HCN(aq) + H2O(l) equilibrium reaction arrow CN −(aq) + H3O+(aq) K = 6.2 10-10 products reactants (b) H2(g) + Cl2(g) equilibrium reaction arrow 2 HCl(g) K = 2.51 104 reactants products
Chemistry
1 answer:
puteri [66]3 years ago
4 0

<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:

  • <u>For a:</u>

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.

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Calculate the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL.
Finger [1]

Considering the definition of dilution, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

<h3>What is diluion</h3>

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

<h3>Molarity of the solution in this case</h3>

In this case, you know:

  • Ci= 4 M
  • Vi= 7 mL
  • Cf= ?
  • Vf= 30 mL

Replacing in the definition of dilution:

4 M× 7 mL= Cf× 30 mL

Solving:

(4 M× 7 mL)÷ 30 mL= Cf

<u><em>Cf= 0.93 mL</em></u>

In summary, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

Learn more about dilution:

brainly.com/question/20113402

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8 0
2 years ago
How many moles are in 8.32 times 1024 molecules of CO2
Svetradugi [14.3K]

Answer:

13.8 mol CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

8.32 × 10²⁴ molecules CO₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u />8.32 \cdot 10^{24} \ molecules \ CO_2(\frac{1 \ mol \ CO_2}{6.022 \cdot 10^{23} \ molecules \ CO_2} ) = 13.816 mol CO₂

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

13.816 mol CO₂ ≈ 13.8 mol CO₂

3 0
3 years ago
How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.48 L against a constant pressure
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p = 0.818 atm = 0.819*101325 Pa = 82985.2 Pa

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Answer: 292.1 J
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Answer:

Sand has a strong non-polar covalent connection, that prevents water to break it down.

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