Answer:
25.1 g is the mass of chlorine in the sample
Explanation:
P . V = n . R . T
We apply the Ideal Gases Law to solve the excersise.
We need to convert the T°C to T°K → 37°C + 273 = 310K
We replace data: 3 atm . 3L = n . 0.082 L.atm /mol.K . 310K
9 atm.L / (0.082 L.atm /mol.K . 310K) = n → 0.354 moles
We convert the moles to mass, to reach the answer
0.345 mol . 70.9g / 1mol = 25.1 g
Answer:
a base
Explanation:
first convert the pOH to pH, that way it will
be easy to decide.
formula:pOH+pH=14
pH=14-4.3=9.7
high pH tells us it's a base
Answer:
19.4 mL Ba(OH)2
Explanation:
H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)
At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.
0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2
Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.
0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl
HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.
0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2