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dimulka [17.4K]
3 years ago
11

Which phase(s) would experience a decrease in the rate of dissolution with an increase in temperature?

Chemistry
2 answers:
lukranit [14]3 years ago
8 0
D. both gases and liquids because as it disolutes the temp increases

Vitek1552 [10]3 years ago
5 0
I woul say that it would have to be gases
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A gas cylinder initially contains 463 L of gas at a pressure of 159 atm. If the final volume of gas is 817 L, what is the final
Agata [3.3K]

Answer:

The final pressure is 90.1 atm.

Explanation:

Assuming constant temperature, we can solve this problem by using <em>Boyle's Law</em>, which states:

  • P₁V₁=P₂V₂

Where in this case:

  • P₁ = 159 atm
  • V₁ = 463 L
  • P₂ = ?
  • V₂ = 817 L

We <u>input the given data</u>:

  • 159 atm * 463 L = P₂ * 817 L

And <u>solve for P₂</u>:

  • P₂ = 90.1 atm

The final pressure is 90.1 atm.

6 0
3 years ago
Ava threw a hamster at her sister the other day she used 12N of force to accelerate the hamster at 8m/s2.what was the mass of th
xenn [34]

<u>We are given:</u>

The force applied on the poor hamster (F) = 12 N

Acceleration of the poor Hamster (a) = 8 m/s²

<u>Solving for the mass of the Poor Hamster:</u>

From newton's second equation of motion, we know that:

F = ma

<em>replacing the given values</em>

12 = 8 * m

m = 12/8 kg

m = 3/2 kg

The poor Hamster weighs 3/2 kg

8 0
2 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
yaroslaw [1]

Answer:

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Explanation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

initially

3.0 atm      0              0

At equilibrium

(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

K_p=\frac{p\times (3p)^3}{(3.0-2p)^2}

=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}

K_p=12.30\approx 12.

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

3 0
3 years ago
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Anvisha [2.4K]

Answer:

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Explanation:

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Explanation:

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