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Molodets [167]
4 years ago
5

A change in which factor will affect the rate of reaction only when gases are involved?

Chemistry
2 answers:
Morgarella [4.7K]4 years ago
6 0
Pressure , can compact the atoms in gases to make them closer and in which case more react-able. Ithink i know its pressure but i dont know how to explain
11Alexandr11 [23.1K]4 years ago
5 0

Answer: Pressure affects the rate of reaction when only gases are involved.

Explanation: For writing the rate of reaction, we use law of mass action, which is:

Rate=\frac{[C]^c[D]^d}{[A]^a[B]^b}

This is the case when all the reactants and products are in solid or liquid states.

But in the case of gases, we replace concentration in the above equation by the partial pressures of reactants and products, so the rate equation becomes:

Rate=\frac{[p_C]^c[p_D]^d}{[p_A]^a[p_B]^b}

So, the factor that effects the rate of reaction when only gases are involved is Pressure.

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you wish to make a 0.161 m nitric acid solution from a stock solution of 3.00 m nitric acid. how much concentrated acid must you
Drupady [299]

From a stock solution of 3.00 m nitric acid, 9.391 ml of stock solution is needed to create a 0.161 m nitric acid solution, which has a total volume of 175 ml of the diluted solution.

A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration. Examples of typical stock solutions in laboratories are nitric acid and hydrochloric acid. These play a critical role in creating the titration-related solution preparations.

We know the formula for dilution type problems

             M1 VI = M2 V 2

Where,

M, = initial molarity

V , = initial Volume

M2 = final molarity

V 2 = final Volume

Hene given -

M, = 3.00 M

VI = ?

M2 = 0.161M

V 2 = 175 ml

Accordingly ' MI V1  = M2 V 2

V1 =(M2*V2)/M1

V1= (0.161M*175ml)/ 3.00M

v1 = 9.391

The required volume of Stock solution is 9.391ml.

Learn more about Stock solution here

brainly.com/question/25256765

#SPJ4

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