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3 years ago

A 3.20-mol sample of gas occupies a volume of 350. mL at 300.0 K. Determine the

Chemistry

1 answer:

nalin [4]3 years ago

4 0

Answer:

$P \approx 225atm$

Explanation:

From the question we are told that:

Moles of sample n=3.20_mol

Volume V=350mL

Temperature T=300k

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

$P=\frac{nRT}{V}$

$P=\frac{3.20*0.08206*300}{350*10^{-3}}$

$P=225.079atm$

$P \approx 225atm$

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Please show all of your work! :)

Paladinen [302]

Answer:

Explanation:

To answer this, we need to use Gay-Lussac's law, which states that:

$\frac{P_1}{T_1}= \frac{P_2}{T_2}$ , where P is pressure and T is temperature

The initial pressure we're given is 4.5 atm (so P1 = 4.5) and the temperature is 45.0°C; however, we need to change Celsius to Kelvins, so add 273 to 45.0: 45.0 + 273 = 318 K (so T1 = 318).

The final pressure is what we want to find, but we do know the final temperature is 3.1°C. Converting this to Kelvins, we get: 3.1 + 273 = 276.1 K, which means T2 = 276.1.

Plug these values in:

$\frac{P_1}{T_1}= \frac{P_2}{T_2}$

$\frac{4.5}{318}= \frac{P_2}{276.1}$

Multiply both sides by 276.1:

$P_2$ ≈ 3.9 atm

The answer is thus A.

3 0

3 years ago

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

5 0

4 years ago

1. Classify rust:

Arisa [49]

1. Rust, or Fe2O3, is a Chemical Compound.

2. Carbon is a Chemical Element found as C on the periodic table.

3. Apple Juice is a Heterogeneous Mixture, as it is uniform throughout. If it wasn't uniform, there would be yucky lumps in your apple juice :)

If I helped you out, I would greatly appreciate if you went through my other answers I gave you and gave them the brainliest answer :)

7 0

3 years ago

an unknown sample required 21.05 mL of 0.02047 M KmnO4 to reach the end point. How many moles of KmnO4 reacted?

Angelina_Jolie [31]

.02047 = x / 0.02105 = 0.0004309mol

6 0

3 years ago

HELP ASAP<br><br> Describe the function of Cytokinesis in the cell cycle.

marta [7]

Cytokinesis ensures that one nucleus ends up in daughter cells during cell division

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3 years ago