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3 years ago

As matter changes, what happens to the energy?

Chemistry

1 answer:

OlgaM077 [116]3 years ago

4 0

Energy, Temperature, and Changes of State

Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid.

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The decomposition of ammonia is: 2 NH3(g) = N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni

hjlf

Answer:

A) = 4.7 × 10⁻⁴atm

Explanation:

Given that,

Kp = 1.5*10³ at 400°C

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

The decomposition reaction is:-

2NH3(g) ↔N2(g) + 3H2(g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 =√ [(pH2)³(pN2)/Kp]

pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm

$K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm$

= 4.7 × 10⁻⁴atm

4 0

3 years ago

Read 2 more answers

Question 8 of 8> 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a s

Nesterboy [21]

Answer:

12.29 M

Explanation:

The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

Now let's calculate the moles of H₂SO₄ that were titrated:

$0.1284 M * 0.1915L*\frac{1molAcid}{2molNaOH}$ = 0.01229 mol H₂SO₄.

Thus, the concentration of the diluted solution is:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

Finally, the concentration of the original acid solution is:

$1.229 M*\frac{250mL}{25mL}$ = 12.29 M

5 0

3 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

3 years ago

When a substance changes from gas to liquid at a temperature below its boiling point, ____ has taken place.

puteri [66]

That would be evaporation.
Hope this helped!! xx

5 0

4 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago