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jek_recluse [69]
3 years ago
8

The making of ice cubes is an endothermic reaction

Chemistry
1 answer:
Rina8888 [55]3 years ago
4 0

Answer:

no

Explanation:

it is not endothermic reaction

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Which reactants would lead to a spontaneous reaction?
sleet_krkn [62]

The reactants that would lead to a spontaneous reaction are Mn2+ and Al3+. Al and Mn  are not because they are stable reactants. Mn and Al3+ are not because Mn is stable. Mn2+ and Al are not because Al is stable.

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Evidence of a chemical change would be
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D- Rusting car fender
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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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These statements describe three different reactions.
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The answer is number 2. That releases massive amounts of radiation and by the way, that is how atomic bombs are made to detonate.
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A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2
Ymorist [56]

Number of moles is defined as the ratio of given  mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = \frac{0.01761 g}{44.01 g/mol}

= 0.0004001 mol

Mass of carbon  = number of moles of carbon dioxide \times molar mass of carbon

= 0.0004001 mol\times 12.011 g/mol

= 0.004806 g

Number of moles of water= \frac{0.00481 g}{18 g/mol}

= 2.672\times 10^{-4}

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = 2\times 2.672\times 10^{-4}

= 5.34\times 10^{-4}

Mass of hydrogen = 5.34\times 10^{-4}\times \times 1.008 g/mol

= 5.34\times 10^{-4} g

Mass of oxygen = 0.001175-(5.38\times 10^{-4}g+0.004806 g)

= 0.006405 g

Number of moles of oxygen = \frac{0.006405 g}{15.999 g/mol}

= 0.000400

Now,

C_{0.0004001}  H_{0.000534}  O_{0.000400}

Divide the smallest number to get the whole number,

C_{\frac{0.0004001}{0.000400}}  H_{\frac{0.000534}{0.000400}}  O_{\frac{0.000400}{0.000400}}

we get,

C_{1}  H_{1.33}  O_{1}

Now, multiply all the subscript by 3 to get the whole number,

C_{3}     H_{4}      O_{3}   (empirical fomula)

Molar mass of the compound  =3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol

= 88.062 g/mol

Divide given molar mass of the compound with the molar mass of the compound.

=\frac{176.1 g/mol}{88.062 g/mol}

= 1.999\simeq 2

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

C_{6}H_{8}O_{6}

Hence, empirical formula is C_{3}H_{4}O_{3} and molecular formula is C_{6}H_{8}O_{6}



8 0
3 years ago
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