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:
CeO₂
Explanation:
Hello!
In this case, since we are given the mass of both cerium and the cerium oxide, we can first compute the moles of cerium and the moles of oxygen as shown below:
Now, we simply divide each moles by 0.03 as the fewest moles in the formula to obtain the simplest formula (empirical formula) of this oxide:
Thus, the formula turns out:
Regards!
Answer:
Check the image file attached
Explanation:
Answer:
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