Molar mass of CH2NH2COOH - 75
Given mass of CH2NH2COOH - 30
Moles of CH2NH2COOH = Given mass/ Molar mass
moles of CH2NH2COOH = 30/75 = 0.4 mol
One mole of CH2NH2COOH contains 32 gram of oxygen
0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen
Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!
the law that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
Answer:
K > 1.
Explanation:
∵ The equilibrium constant K = [products]/[reactants].
Since, [products] > [reactants].
<em>∴ The equilibrium constant K > 1.</em>
Answer:
Pp O2 = 82.944 KPa
Explanation:
heliox tank:
∴ %wt He = 32%
∴ %wt O2 = 68%
∴ Pt = 395 KPa
⇒ Pp O2 = ?
assuming a mix of ideal gases at the temperature and volumen of the mix:
∴ Pi = RTni/V
∴ Pt = RTnt/V
⇒ Pi/Pt = ni/nt = Xi
⇒ Pi = (Xi)*(Pt)
∴ Xi: molar fraction (ni/nt)
⇒ 0.68 = mass O2/mass mix
assuming mass mix = 100 g
⇒ mass O2 = 68 g
∴ molar mass O2 = 32 g/mol
⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2
⇒ mass He = 32 g
∴ molar mass He = 4.0026 g/mol
⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He
⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles
molar fraction O2:
⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099
⇒ Pp O2 = (X O2)(Pt)
⇒ Pp O2 = (0.2099)(395 KPa)
⇒ Pp O2 = 82.944 KPa
