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Lubov Fominskaja [6]
2 years ago
11

A compound has a percent composition of 15.24% sodium (molar mass = 22.99

Chemistry
1 answer:
RUDIKE [14]2 years ago
8 0

The empirical formula of the compound obtained from the question given is NaBrO₃

<h3>Data obtained from the question </h3>
  • Sodium (Na) = 15.24%
  • Bromine (Br) = 52.95%
  • Oxygen (O) = 31.81%
  • Empirical formula =?

<h3>How to determine the empirical formula </h3>

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Na = 15.24 / 22.99 = 0.663

Br = 52.95 / 79.90 = 0.663

O = 31.81 / 16 = 1.988

Divide by the smallest

Na = 0.663 / 0.663 = 1

Br = 0.663 / 0.663 = 1

O = 1.988 / 0.663 = 3

Thus, the empirical formula of the compound is NaBrO₃

Learn more about empirical formula:

brainly.com/question/24297883

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Hey there!:

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Answer:

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Explanation:

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3 years ago
The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::
Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

<u>v = k[NO]² [O₂]</u>

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

<u>M⁻² s⁻¹ = k </u>

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = <u>7.11 x 10³ M⁻² s⁻¹ </u>

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = <u>0.8 M/s</u>

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = <u>0.4 M/s</u>

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

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