Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
This question is incomplete. Luckily, I found the same problem which is shown in the attached picture. To answer the question, we must know how the size and charge affect the lattice energy. The answer is: lattice energy increases with the increasing charge of the ions, and decreasing radius of the atoms.
<em>Therefore, the ranking would be: A < B < C</em>.
Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole
20 grams of borax contains (20.0g) / (201 g mol -1) =0.10 mol of borax.
Therefore 0.40 mol of borax
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
Mass defect Δ m = 236.052595 - 235.854000 = 0.198 amu
Reaction energy released Q = Δ m * 931.5
= 0.198 * 931.5 = 185 MeV
