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3 years ago

How many molecules are in 1.95 moles of water?

Chemistry

2 answers:

Licemer1 [7]3 years ago

8 0

Answer: 1.1739*10²⁴ molecules

Explanation:

Using Avogadro's constant,

1 mole of any substance = 6.02*10²³ molecule/atom or particles.

Therefore 1.95 mole of water contains;

1.95 * 6.024*10²³ molecules = 1.1739*10²⁴ molecules.

OLEGan [10]3 years ago

3 0

Water has a molar mass of 18.015 g/mol . This means that one mole of water molecules has a mass of 18.015 g . So, to sum this up, 6.022⋅1023 molecules of water will amount to 1 mole of water, which in turn will have a mass of 18.015 g . 2.7144moles H2O ⋅6.022⋅1023molec.1mole H2O =1.635⋅1024molec.

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3 years ago

Read 2 more answers

Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3

Margarita [4]

Answer: The rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

$2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O$

The intermediate reaction of the mechanism follows:

Step 1: $H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}$

Step 2: $H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}$

Step 3: $HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[H_3O_2^+][Br^-]$ ......(1)

As, $[H_3O_2^+]$ is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for $[H_3O_2^+]$ from step 1, we get:

$K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}$

$[H_3O_2^+]=K[H^+][H_2O_2]$

Putting the value of $[H_3O_2^+]$ in equation 1, we get:

$\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]$

Hence, the rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

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