Answer: 94.07%
Explanation:
Percentage yield can be calculated by the formula
%yield = Experimental yield/Theoretical yield x100
Experimental yield = 7.93g
Theoretical yield = 8.43
%yield = Experimental yield/Theoretical yield x100
%yield = 7.93/8.43 x 100 = 94.07%
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Heat required in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. It is
expressed as follows:<span>
Heat = mC (T2-T1)
Heat = 10.0 g (4.18 J/g-C ) ( 6.0 C )
<span>Heat = 250.8 J</span></span>
The question does not provide the equation
Answer:-
72.89 grams
Explanation:-
The balanced chemical equation for this reaction is
CuSO4 + Fe --> FeSO4 + Cu
Molecular weight of CuSO4 = 63.55 x 1 + 32 x 1 + 16 x 4
= 159.55 gram
Atomic weight of Cu = 63.55 gram.
According to the balanced chemical equation
1 CuSO4 gives 1 Cu
∴159.55 gram of CuSO4 would give 63.55 gram of Cu.
183 gram of CuSO4 would give 63.55 x 183 / 159.55
= 72.89 grams of Cu
N is as electronegative as Cl is. So Cl doesnt react with O, N