All categories

2 years ago

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade

Chemistry

1 answer:

egoroff_w [7]2 years ago

5 0

Answer:

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

Explanation:

Assuming Pressure is constant.

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

$\frac{300}{273} = \frac{V_2}{288}$

$V_2 = \frac{288*300}{273} \\V_2 = 316.5$

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

You might be interested in

A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

rodikova [14]

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

6 0

3 years ago

How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5

NeTakaya

Answer:

Explanation:

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

C) 2 Is the correct answer, I took the test and it was correct.

5 0

3 years ago

Read 2 more answers

Is the oxygen you breathe brand new

PilotLPTM [1.2K]

Answer:

Yes because every plant around you keeps giving new oxygen.

3 0

3 years ago

Identify the relationship between <2 and <4*

dexar [7]

I don’t understand your questions can you elaborate

6 0

3 years ago

What draws water back to earth

spayn [35]

Gravity is the force that draws water back to the earth in the forms of rain, sleet and, snow.

7 0

3 years ago

Read 2 more answers

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade​

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade