Answer:
- Part a) 0.0104 moles copper(II) nitrate.
i) 0.0418 mole Cu
ii) 0.0209 mol Ag NO₃
Explanation:
<u>1) Balanced chemical reaction (single replacement):</u>
In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)
- Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag
<u>2) Mole ratio: </u>
- 1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag
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<u>3) Moles of Ag</u>
- n = mass in grams / atomic mass
- atomic mass of Ag: 107.868 g/mol
- n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag
<u>4) Moles of copper(II) nitrate:</u>
- Set the proportion using the mole ratio:
- 2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x
- Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ = 0.0104 moles Cu(NO₃)₂
That is the answer of part a: 0.0104 moles copper(II) nitrate.
<u>5) Moles of each reactant</u>
i) Cu:
- Set a proportion using the theoretical mole ratio
1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag
- Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu
ii) Ag NO₃
- Set a proportion using the teoretical mole ratio
2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag
- Solve for x: x = 0.0209 mol Ag NO₃
Answer:
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Explanation:
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radioactive element will be left at 2 p.m. The correct answer between
all the choices given is the second choice or letter B. I am hoping that this
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