All categories

3 years ago

A gas occupies 40.0 mL at 150 K. What volume does it occupy at 300 K,

Chemistry

1 answer:

nlexa [21]3 years ago

6 0

Answer:

.08 L or 80 ml

Explanation:

Use the equation V/t = V/t.

.04L / 150K = V / 300K

.04 / 150 * 300 = V

.08 L or 80 ml

You might be interested in

Consider the balanced equation below

Svet_ta [14]

PCl₃ + Cl₂ = PCl₅

1:1

3 0

3 years ago

Read 2 more answers

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

A) Calculate the pH of a 2.0x10-4 M solution of aniline hydrochoride, C6H5NH3Cl.

igor_vitrenko [27]

Given the concentration of aniline hydrochloride is $2.0 *10^{-4} M$

Aniline hydrochloride is the conjugate acid of aniline a weak base.

pH can be calculated from $pK_{a}$ anilinium ion the conjugate acid of aniline.

8 0

3 years ago

Read 2 more answers

A certain indicator has a color chart that shows that when the indicator is yellow, the solution is acidic. What is true about a

bija089 [108]

It’s acidic?
It tastes sour
It has at least one hydrogen molecule

4 0

3 years ago

How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?

boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0

3 years ago