Answer:12 mol
Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).
6.7mol per 1.3L = 6.7/1.3 mol/L
so in 2.33L = 6.7*2.33/1.3 = 12 mol
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
Alkali metals.
Elements found in group 1 of the periodic table.
Answer:
Volume = 44.96L
Explanation:
p = 427mmHg = 0.56atm
R = 0.08206
T 6°C = 279k
V = ?
n = 1.10 mol
using ideal gas equation
pv=nRT
0.56 x V = 1.10 x 0.08206 x 279
0.56 x V = 25.18
V = 25.18/0.56
V = 44.96L