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Dafna1 [17]
3 years ago
5

2. What is the molarity of 100 mL of a 3.0% H2O2 (mass/volume) solution? What is the molarity of 100 mL of a 2.25% H2O2 solution

?
Chemistry
1 answer:
KatRina [158]3 years ago
7 0

The 3% mass/volume H₂O₂ means 3 g of H₂O₂ in 100 ml of water.

Now, Molarity (M) = No. of moles of H₂O₂ / Volume of solution in liter

No. of moles of H₂O₂ = Mass / Molar mass = 3 g / 34 g/mol = 0.088 mol

So, molarity = 0.088 × 1000 ml / 100 ml = 0.88 M

In case of 2.25 % H₂O₂,

No of moles = 2.25 g / 34 g/mol = 0.066 mol

Molarity = 0.066 mol / 0.100 L = 0.66 M.

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An = Aoe^-kt

We use this equation for the solution. We do as follows:

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If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
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This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

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Concentration at equilibrium:

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Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

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When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

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