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lubasha [3.4K]
3 years ago
11

When you decrease the volume of a gas, what happens to the temperature?

Chemistry
1 answer:
ale4655 [162]3 years ago
4 0

the temperature is decreased

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umka21 [38]
Covalent compounds are formed when non metals bond together.
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3 years ago
Explain why a solution that is 1.3 M HF and 1.3 mM KF is not a good buffer. HF is a strong acid and cannot be used in a buffer s
zavuch27 [327]

Answer:

The ratio of acid to conjugate base is outside the buffer range of 10:1.

Explanation:

The Henderson-Hasselbalch equation for a buffer is

\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}

A buffer should have

\dfrac{1}{10} \leq \dfrac{\text{[A$^{-}]$}}{\text{[HA]}} \leq \dfrac{10}{1}

For a solution that is 1.3 mol·L⁻¹ in HF and 1.3 mmol·L⁻¹ in KF, the ratio is

\dfrac{1.3 \times 10^{-3} }{1.3} = \dfrac{1}{1000}

The ratio of acid to conjugate base is 1000:1, which is outside the range of 10:1.

A is wrong. NF is a weak acid.

C is wrong. The two species are a conjugate acid-base pair.

D is wrong. Salts of Group 1 metals are soluble.

4 0
3 years ago
At a certain temperature, the K p Kp for the decomposition of H 2 S H2S is 0.834 . 0.834. H 2 S ( g ) − ⇀ ↽ − H 2 ( g ) + S ( g
stira [4]

Answer:

Total pressure at equilibrium is 0.2798atm.

Explanation:

For the reaction:

H₂S(g) ⇄ H₂(g) + S(g)

Kp is defined as:

Kp = \frac{P_{H_{2}}*P_S}{P_{H_{2}S}} = 0.834

If initial pressure of H₂S is 0.150 atm, equilibrium pressures are:

H₂S(g): 0.150atm - x

H₂(g): x

S(g): x

Replacing in Kp:

\frac{X*X}{0.150atm-X} = 0.834

X² = 0.1251 - 0.834X

X² +  0.834X - 0.1251 = 0

Solving for X:

X = -0.964 → False solution: There is no negative pressures

X = 0.1298

Thus, pressures are:

H₂S(g): 0.150atm - 0.1298atm = <em>0.0202atm</em>

H₂(g): <em>0.1298atm</em>

S(g): <em>0.1298atm</em>

Thus, total pressure in the container at equilibrium is:

0.0202atm + 0.1298atm + 0.1298atm = <em>0.2798atm</em>

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3 years ago
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