Answer: A little bit confused can you explain what I have to do
Explanation:
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!
Answer: 4.12
Explanation:
we know that the given mol is 8.23 mol and they are 2NaI and I2 so we will write the equation like this.
8.23mol NaI x 1mol of I2 ÷ 2molNaI = 4.115≅ 4.12 mol of I2
we placed NaI at the bottom to cancel out with the 8.23 mol of NaI
Explanation:
Below is an attachment containing the solution.