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Elis [28]
3 years ago
6

Which word identifies the shaking that results from movement under Earth's surface?

Chemistry
2 answers:
djyliett [7]3 years ago
8 0
The anser is plant that what i think
Valentin [98]3 years ago
8 0

Answer:

A. Earthquake

Explanation:

took the test :)

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How is the oxidation state of a transition metal determined from the chemical formula ?
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Answer:

Explanation:

In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.

We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.

The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.

To ascertain the oxidation state, we have to comply with some rules:

  • The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
  • The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to  the charge on the ion.

For example, let us find the oxidation state of Cr in Cr₂O₇²⁻

This would be:  2x + 7(-2) = -2

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We see that the oxidation number of Cr, a transition metal in the given ion is +6.

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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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