A sample of oxygen gas has a volume of 36.7 L at 145 kPa and 65. ̊C. What volume will the sample have at STP?

1 answer:

8 0

Data Given:

Initial Volume = V₁ = 36.7 L

Initial Pressure = P₁ = 145 kPa

Initial Temperature = T₁ = 65 °C + 273 = 338 K

Final Volume = V₂ = ?

Final Pressure = P₂ = 101.325 kPa (Standard Pressure)

Final Temperature = T₂ = 273 K (Standard Temperature)

Formula used:
As number of moles are constant, so Ideal Gas equation in following form is used,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / T₁ P₂
Putting Values,

V₂ = (145 kPa × 36.7 L × 338 K) ÷ (273 K × 101.325 kPa)

V₂ = 1798667 ÷ 27661.25

V₂ = 65.02 L

You might be interested in

• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo

garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c) 0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

1 mole 3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

$H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O$

10 ml 17.50 ml

(x) M 0.200 M

Molarity = $\frac{0.2*17.5}{1000}$