Answer:
are you rich in points brother ? or sister ?
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
<em></em>
Mol sulfuric acid = 19 g * (1 mol) / (98.1 g) = 0.19367 mol
mol H2O = 0.19367 mol H2SO4 * (2 H2O) / (1 H2SO4)
= 0.387359 mol H2O
grams H2O = 0.387359 mol H2O * (18 g)/(1 mol)
= 6.97 g
The answer is 7.0 grams of water
Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
Answer:
5)0.53liters
6)1.634.6°c
Explanation:
5) v1=25liters, t1=12000°c. v2=?, t2=250°c
v2=(250×25)/12000
v2=6250/12000
v2=0.52liters
6)v1=130,v2=85, t1=2500, t2=?
t2=(85×2500)/130
t2=212500/130
t2= 1634.6°c
