The first choice might be correct. I'm sincerely sorry if I am mistaken!
FIRST OPTION (A)(1)
Elemental zinc replaces copper
Explanation:
During this reaction, the zinc added to the copper sulfate solution has replaced the copper present in the compound.
This process is known as a single displacement reaction.
Zn + CuSO₄ → ZnSO₄ + Cu
This represents the reaction in the recycling process.
The reaction is influenced by the elements' positions within the reactivity series of metals.
- In a single displacement reaction, an element higher in the reactivity series displaces one that is lower.
- Zinc ranks above copper in this series, allowing it to react with sulfate.
- Consequently, copper is pushed out as a solid product in the solution.
- Elements positioned higher in this series exhibit greater reactivity
learn more:
Chemical reaction
Let's assume that the compound formula is as follows: Experiment 1: 1.00 g of the compound yields 1.95 g of AgCl. The molar mass of AgCl is 143.32 g/mol. Thus, the moles of AgCl for 1.95g are: The moles of Cl also equal 0.0136, considering that 1 mole of AgCl corresponds to 1 mole of Cl. Experiment 2: 1.00 g of the compound results in 0.900 g of CO2 and 0.735 g of H2O. The molar mass of CO2 is 44 g/mol, and for H2O, it's 18 g/mol. Therefore, the moles of C come to 0.0205 and the moles of H stand at 0.0816 (which is 2 times the moles of H2O). Now, from the provided details, it's derived that in 1.00 g of the compound, there are 0.0136 moles of Cl, 0.0205 moles of C, and 0.0816 moles of H. In terms of mass: Mass of Cl = 0.0136 * 35.5 = 0.4828 g. Mass of C = 0.0205 * 12 = 0.246 g. Mass of H = 0.0816 * 1 = 0.0816 g. Total mass = 0.4828 + 0.246 + 0.0816 + mass of N. Given that 1.00 = 0.8104 + Mass of N, it follows that Mass of N = 0.1896. Thus, upon dividing all moles by the smallest value, we find Cl = 0.0136 / 0.0135 = 1.0007; C = 0.0205 / 0.0135 = 1.52; H = 0.0816 / 0.0135 = 6.04; N = 0.0135 / 0.0135 = 1. Multiplying by 2 allows us to reach integer values: Cl = 2, C = 3, H = 12, N = 2.
M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL
From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
Answer:
The integer value of x in the hydrate is 10.
Explanation:
Molar concentration of the solution = 0.0366 M
Volume of the solution = 5.00 L
Moles of hydrated sodium carbonate = n
Weight of hydrated sodium carbonate = n = 52.2 g
Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol
By solving for x, we arrive at:
x = 9.95, approximating to 10
The integer x in the hydrate equals 10.
