Answer:
I think it is B
Explanation:
because the others don't look right
Answer:
2Al+1.5O2→Al2O3
Thus, 2 mol of Al combine with 1.5 mol of oxygen to form 1 mol of Al2O3.
2 mol of Al corresponds to 2×27=54 g.
Thus, the weight of Al used in the reaction is 54 g.
Use the state equation for ideal gases: pV = nRT
Data:
V = 88.89 liter
n = 17 mol
T = 67 + 273.15 = 340.15 K
R = 0.0821 atm * liter / (K*mol)
=> p = nRT / V = 17 mol * 0.0821 (atm*liter / K*mol) * 340.15 K / 88.89 liter
p = 5.34 atm
Answer: p = 5.34 atm
Answer:
(a) weight percent of Cu = 44.59%
(b) weight percent of Zn = 53.49%
(c) weight percent of Pb = 1.91%
Explanation:
Given: Alloy contains- Cu=93.1 g, Zn=111.7 g, Pb=4.0 g
Therefore, the total mass of the alloy (m) = mass of Cu (m₁) + mass of Zn (m₂)+ mass of Pb (m₃)
m= 93.1 g + 111.7 g + 4.0 g = 208.8 g
(a) weight percent of Cu = (m₁ ÷ m)× 100% = (93.1 g ÷ 208.8 g)× 100% =44.59%
(b) weight percent of Zn = (m₂ ÷ m)× 100% = (111.7 g ÷ 208.8 g)× 100% =53.49%
(c) weight percent of Pb = (m₃ ÷ m)× 100% = (4.0 g ÷ 208.8 g)× 100% =1.91%
