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
<span>1. Seeing broader picture: determining gray areas, overlaps, and exclusions, biases, the primary focus.
2. Bias. Sources may have hidden agenda, personal discrimination conscience or unconscious and a limited view of the short term, long term effects.</span>
Do the atomic mass times the number of atoms you have. That'll give you your answer. Hope it helps!
175.8 g NaCl to moles:
(175.8 g)/(58.44 g/mol) = 3.008 mol NaCl
Molarity = moles of solute/volume of solution in liters
(3.008 mol NaCl)/(1.5 L) = 2.0 M.
The molarity of this solution would be 2.0 M.
Answer:
pH is 7.60
Explanation:
Let's think the equations:
HClO + H₂O ⇄ ClO⁻ + H₃O⁺
As every weak acid, we make an equilbrium
The salt is dissociated in solution
NaClO → Na⁺ + ClO⁻
HClO + H₂O ⇄ ClO⁻ + H₃O⁺
Initially 0.3m 0.35m
We have the moles of acid, and the moles of conjugate base.
Reacts x X X
Some amount has reacted, so I obtained (in equilibrium) the moles of base + that amount, and the same amount for H₃O⁺ (ratio is 1:1)
HClO + H₂O ⇄ ClO⁻ + H₃O⁺
0.3 - x 0.35 + x x
Let's make the expression for Ka
Ka = [ClO⁻] . [H₃O⁺] / [HClO]
(we don't add water, because it is included in Ka)
2.9x10⁻⁸ = (0.35+x).x / (0.3-x)
Ka is in order of 10⁻⁸, I can assume that 0.3-x is 0.3 and 0.35 +x =0.3
2.9x10⁻⁸ = (0.35)x / (0.3)
(2.9x10⁻⁸ . 0.3) /0.35 = x
2.48x10⁻⁸ = x
This is [H₃O⁺]
For pH = - log [H₃O⁺]
pH = 7.60
