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3 years ago

Summarize the relationship between volume, temperature and pressure (use words like “inversely

Chemistry

1 answer:

Alinara [238K]3 years ago

4 0

Answer:

P and V: inversely proportional

P and T: directly proportional

V and T: inversely proportional

Explanation:

For pressure and volume, as the volume goes up, meaning the container gets bigger, the pressure would go down. There would be more room in the container, so there would be less collisions between the molecules themselves and between the molecules and the container. This makes them inversely proportional.

For pressure and temperature, as the pressure goes up, there are more collisions, so the particles move faster. Temperature is the speed of the particles, so, since both pressure and temperature would go up at the same time, they are directly proportional.

For volume and temperature, this is similar to the PV relationship. As volume increases, there are less collisions between the particles. This means that the particles are going to move slower. Therefore, as volume goes up, temperature goes down, so they are inversely proportional.

Sorry this is super long, but I hope it fully explains the question for you! ☺

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Write the formula for the conjugate base for the following acids. Thank you

Llana [10]

Answer:

See the answer and explanation below , please.

Explanation:

A conjugate base is defined as that formed after an acid donates its proton.

For each article, a continuation of the conjugate bases (highlighted in bold), for dissociation in water:

a) HF + H20 --> F- + H30+

b) H20+ H20 --> OH- + H30+

C)H2PO3- + H20--> HPO3 2- + H30+

d) HSO4- + H20 --> SO4 2- + H30+

E) HCL02 + H20 --> CLO02 - + H30+

7 0

3 years ago

Read 2 more answers

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

A) ambos pertencem ao 5o período da Tabela Periódica;

Reptile [31]

Answer:

Explanation:

8 0

3 years ago

1. Do you think that the lightbulb and the Moon spheres are "to scale" compared to the real

Tems11 [23]

Answer:

Yes

Explanation: Had a question like this and I said yes and got it right

3 0

2 years ago

4. A student is doing experiments with CO2(q) . Originally a sample of the gas is in a rigid container at 299 Kand 0.70 atm. The

marta [7]

Answer:

b. 0,99atm

c. Answer is in the explanation

d. Answer is in the explanation

Explanation:

b. Using Gay-Lussac's law:

P₁T₂ = P₂T₁

P₁: 0,70 atm; T₂: 425K; P₂: ??; T₁: 299K

0,70atm×425K / 299K = 0,99 atm

c. Using kinetic molecular theory, the increasing of temperature increases the kinetic energy of gas particles and if kinetic energy increases, the pressure increases. That means the increasing of temperature increases the pressure in the system.

d. Now, the increases in kinetic energy of gases increase the collisions betwen particles. As these intermolecular forces that are not taken into account in ideal gas law, the observed pressure will be different to the pressure predicted by ideal gas law.

I hope it helps!

8 0

3 years ago