Answer:
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Explanation:
Mass of silver to be precipitated on ecah spoon = 0.500 g
Number of silver spoons = 250
Total mass of silver = 250 × 0.500 g = 125 g
![Moles (n)=Molarity(M)\times Volume (L)](https://tex.z-dn.net/?f=Moles%20%28n%29%3DMolarity%28M%29%5Ctimes%20Volume%20%28L%29)
Moles of AgCN = n = ![\frac{125 g}{134 g/mol}=0.9328 mol](https://tex.z-dn.net/?f=%5Cfrac%7B125%20g%7D%7B134%20g%2Fmol%7D%3D0.9328%20mol)
Volume of AgCN solution =V
Molarity of the AgCN = 2.50 M
![V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B0.9328%20mol%7D%7B2.50%20M%7D%3D0.3731%20L%3D373.1%20mL)
(1 L = 1000 mL)
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
78 protons are found in platinum(pt).
Answer:
4.5 moles of lithium sulfate are produced.
Explanation:
Given data:
Number of moles of lead sulfate = 2.25 mol
Number of moles of lithium nitrate = 9.62 mol
Number of moles of lithium sulfate = ?
Solution:
Chemical equation:
Pb(SO₄)₂ + 4LiNO₃ → Pb(NO₃)₄ + 2Li₂SO₄
Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.
Pb(SO₄)₂ : Li₂SO₄
1 : 2
2.25 : 2/1×2.25 = 4.5 mol
LiNO₃ : Li₂SO₄
4 : 2
9.62 : 2/4×9.62 = 4.81 mol
Pb(SO₄)₂ produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of Li₂SO₄.
Density = mass/volume = 800/200 =4 g/cm^3