Answer: 72.41% and 26.90% respectively.
Explanation:
At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.
We can calculate recovery as:
So the answer to the first question is 72.41%.
For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.
So the answer to the second question is 26.90%.
Answer:
My answer im guessing is "d"
Explanation:
Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
The Actual Yield is given in the question as 21.2 g of NaCl. However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.
Balanced Equation: CuCl + NaNO₃ → NaCl + CuNO₃
Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
= 31.0 g ÷ (63.5 + 35.5)g/mol
= 0.31 mol
the mole ratio of CuCl to NaCl is 1 : 1,
∴ if moles of CuCl = 0.31 mol,
then moles of NaCl = 0.31 mol
Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
= 0.31 mol × (23 + 35.5) g/mol
= 18.32 g
⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.
Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100
⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100
= 115.7 %
NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.
The electron configuration
1
s
2
2
s
2
2
p
6
3
s
2
3
p
2
is the element Silicon.
The key to deciphering this is to look at the last bit of information of the electron configuration
3
p
2
.
The '3' informs us that the element is in the 3rd Energy Level or row of the periodic table. The 'p' tells us that the element is found in the p-block which are all of the Groups to the right of the transition metals, columns 13-18. The superscript '2' tells us that the element is found in the 2nd column of the p-block Group 14.
