Which of the following pairs of elements are chemically most similar to each other?

What is the natural rate of nitrogen fixation in Earth’s ecosystems? What is the natural rate of nitrogen fixation in Earth’s ec

quester [9]

Answer:

100 teragrams of nitrogen per year

Explanation:

Nitrogen fixation in Earth's ecosystems is defined as a process where by nitrogen in air is transformed into ammonia or other related nitrogenous compounds. Generally, atmospheric nitrogen is referred to as molecular dinitrogen and it is a nonreactive compound that is metabolically useless to all but a few microorganisms. This process is vital to life due to the fact that inorganic nitrogen compounds are needed for the biosynthesis of amino acids, protein, and all other nitrogen-containing organic compounds. Thus, the natural rate of nitrogen fixation in Earth's ecosystems is 100 tetragrams of nitrogen per year.

7 0

3 years ago

When the pressure that a gas exerts on a sealed container changes from 22.5 psi to ? psi, the temperature changes from 110 degre

natta225 [31]

Using Gay-Lussac's Law, pressure is proportional to (absolute) temperature in Kelvin. We first convert the temperature values to Kelvin: 110 C = 383.15 K, while 65 C = 338.15 K.
P1/T1 = P2/T2
22.5/383.15 = P2/338.15
P2 = 19.9 psi

8 0

3 years ago

Which sentence from A Girl from Yamhill best illustrates the author's use of sensory language? (A) I was not afraid and did not

oksano4ka [1.4K]

Answer:

D

Explanation:

This sentence has the most sensory details or details giving more description of the 5 senses.

Hope this helps :)

6 0

3 years ago

Read 2 more answers

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:

v = k[NO]² [O₂]

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

M⁻² s⁻¹ = k

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 = 7.11 x 10³ M⁻² s⁻¹

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 = 0.8 M/s

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 = 0.4 M/s

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).

3 0

3 years ago

A solution is made by dissolving 10.7 g of magnesium sulfate, MgSO4, in enough water to make exactly 100 mL of solution. Calcula

adell [148]

Answer:

[MgSO₄] = 890 mM/L

Explanation:

In order to determine molarity we need to determine the moles of solute that are in 1L of solution.

Solute: MgSO₄ (10.7 g)

Solvent: water

Solution: 100 mL as volume. (100 mL . 1L / 1000mL) = 0.1L

We convert the solute's mass to moles → 10.7 g / 120.36 g/mol = 0.089 moles

Molarity (mol/L) → 0.089 mol/0.1L = 0.89 M

In order to calculate M to mM/L, we make this conversion:

0.89 mol . 1000 mmoles/ 1 mol = 890 mmoles

4 0

3 years ago