0.04350179862
Explanation:
Make sure to check for sig figs though. Basically, you have 2.54 mL and you multiply that by the 0.789 g/mL and mL cancels out and you are left with 2.00406g. There are 46.06844 grams per mol of ethanol, so to cancel out grams we multiply by 1/46.06844. And then we are left with 0.04350179862 mol
Answer:
21.3dm³
Explanation:
Given parameters:
Initial volume of gas = 3dm³
Initial pressure = 101kPa
Final pressure = 14.2kPa
Unknown
Final volume = ?
Solution:
To solve this problem, we use the Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".
Mathematically;
P₁ V₁ = P₂ V₂
P and V are pressure and Volume
1 and 2 are the initial and final states
Now insert the parameters and solve;
101 x 3 = 14.2 x V₂
V₂ = 21.3dm³
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
<u />
<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
