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:
The masses of the reactants and products are equal.
Explanation:
Hope this helps.
The answer is 19.9 grams cadmium.
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
heat lost by cadmium = heat gained by the water
-qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
-qcadmium = qwater
-(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
mcadmium = 19.9 grams
Answer:
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
Explanation:
The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.
Where:
= Vapor pressure of pure solvent
= Vapor pressure of the solution
= Number of moles of solvent
= Number of moles of solute
Mass of 0.05499 moles of estrogen :
= 0.05499 mol × 272.4 g/mol = 14.9802 g
14.9802 grams of estrogen must be added to 216.7 grams of benzene.
