I agree with you. At

**STP**, the volume of one mole of gas is 22.4 liters. To calculate the...2 Answers · Science & Mathematics · 15/11/2019

...ideal gas law .. (2) what does S and T and P stand for in

**STP**.. (3) is it easier to write**STP**or T = 273.15K...1 Answers · Science & Mathematics · 28/05/2020

Since 1982 the definition of

**STP**has been 273.15 K and exactly 10^5 Pa. (5.0 mL) x (100 kPa / 101.325 kPa) x (273.15 K / (273.15 + 22)K) = 4.5668 mL = 4.6 mL1 Answers · Science & Mathematics · 08/02/2020

...mole H2/ 2 mole Al) x 0.0815 = 0.122 mol 1 mole of any ideal gas at

**STP**occupies 22.4 L so H2 occupies 22.4 x 0.122 = 2...2 Answers · Science & Mathematics · 21/04/2020

72.1 L O2 / 22.4 L/mol = 3.22 mol O2 108 g Fe / 55.845 g/mol = 1.93 mol Fe 1.93 mol Fe X (3 mol O2/4 mol Fe) = 1.45 mol O2 to react completely with available Fe. So, Fe is the limiting reactant 1.93 mol...

3 Answers · Science & Mathematics · 26/05/2020

(3.17 g/L) / (22.4 L/mol) = 71.008 g/mol The common gas with a molar mass like that is chlorine (Cl2).

2 Answers · Science & Mathematics · 27/11/2019

Exact question: https://answers.yahoo.com/question/index?qid=20200522022256AA4g9c5

3 Answers · Science & Mathematics · 22/05/2020

2 H2O2 → 2 H2O + O2 (30 g H2O2) / (34.01468 g H2O2/mol) x (1 mol O2 / 2 mol H2O) x (22.414 L/mol) = 9.88 L So answer c.

1 Answers · Science & Mathematics · 08/06/2020

P₁V₁/T₁ = P₂V₂/T₂ (150 kPa x 1 atm / 101.325 kPa))(6.85 L) / (35.0°C + 273.15)K = (1 atm)V₂/(273.15 k) V₂ = (150 kPa x 1 atm / 101.325 kPa)(6.85 L)(273.15 k)/[(1 atm)(35.0°C + 273.15)K...

1 Answers · Science & Mathematics · 28/05/2020

1 mole = 22.4 liters 22.4 liters × .005 = .112 liters = 112 cm³ ³√112 = 4.82 cm edge length.

1 Answers · Science & Mathematics · 28/05/2020