Number of moles of oxygen = x
number of moles of nitrogen = y
x = 2y
initial pressure, p1 = 0.8 atm
final pressure, p2 = 1.10 atm
At constant volume and temperature p1 / n1 = p2 / n2
=> p1 / p2 = n1 / n2
n1 = x + y = 2y + y = 3y
n2 = 0.2 + 3y
=> p1 / p2 = 3y / (0.2 + 3y)
=> 0.8 / 1.10 = 3y / (0.2 + 3y)
=> 0.8 (0.2 + 3y) = 1.10 (3y)
0.16 + 2.4y = 3.3y
=> 3.3y - 2.4y = 0.16
=> 0.9y = 0.16
=> y = 0.16 / 0.9
=. x = 2*0.16/0.9 = 0.356
Answer: 0.356 moles O2
Answer:
.371 mole of NaCl
Explanation:
Na Cl Mole weight = 22.989 + 35.45 = 58.439 g/mole
21.7 g / 58.439 g/mole = .371 mole
Realize that pH + pOH = 14
so, 9 + pOH = 14 -> pOH = 5
pOH = -log[OH-]
5 = -log[OH-]
plug it into a calculator and you get 1.0 x 10^-5
alternatively, use [OH-] = 10^-pOH to get the same answer
[OH-] = 1.0 x 10^-5
I also think it’s B but not quite sure
