All categories

3 years ago

What is the volume occupied by 0.103 mol of helium gas at a pressure of 0.95 atm and a temperature of 303 K ?

Chemistry

1 answer:

Finger [1]3 years ago

7 0

The pressure of He gas, P = 0.95 atm

Temperature, T = 303 K

Gas constant, R = 0.0821 L.atm.K⁻¹mol⁻¹

Number of moles of He gas, n = 0.103 mol

According to an ideal gas equation, PV = nRT

V = nRT/P

= (0.103 mol) (0.0821 L.atm.K⁻¹mol⁻¹) (303 K) / 0.95 atm

= 2.7 L

Thus, the volume of the He gas is 2.7 L.

You might be interested in

You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution

e-lub [12.9K]

Answer:

0.85 Molar Na2O

Explanation:

Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

(10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.

Molar is a measure of concentration. It is defined as moles/liter. A 1 M solution contains 1 mole of solute per liter of solvent. [200 ml water = 0.2 Liters water.]

In this case, we have 0.161 moles Na2O in 0.200 L of solvent.

(0.161 moles Na2O)/(0.200 L) = 0.85 Molar Na2O

8 0

1 year ago

Who wants to make some Chemistry????... UwU

Alik [6]

Answer:

what kind of chemistry is it going to be?

8 0

2 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$