Answer:
the percentage composition of mass of nitrogen in NH OH is 42.86 %
N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
(R = 22.4/273 = 0.082)
= (5.49 x 22.03)/(0.082 x 361.78) = ?
Put it into the calculator. It's hard to do that on a mobile phone.
Answer:
a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂
b) Ni(OH)₂
c) KOH
d) 0.927 g
e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M
Explanation:
a) The equation is:
2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)
b) The precipitate formed is Ni(OH)₂
c) The limiting reactant is:


From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:
Hence, the limiting reactant is KOH.
d) The mass of the precipitate formed is:
e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:


I hope it helps you!