3.
1 answer:
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Explanation:
Molarity is defined as number of moles per liter of solution.
Mathematically, molarity =
It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.
molarity =
0.0800 M =
no. of moles = 1.6 mol
Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.
No. of moles =
mass in grams =
=
= 399.488 g
Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.
Answer:
The percent yield of chloro-ethane in the reaction is 82.98%.
Explanation:
Moles of ethane =
Moles of chlorine gases =
As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.
This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.
According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.
Then 9.1549 moles of chlorien gas will give:
of chloro-ethane
Mass of 9.1549 moles of chloro-ethane:
9.1549 mol × 64.5 g/mol = 590.4910 g
Theoretical yield of chloro-ethane: 590.4910 g
Given experimental yield of chloro-ethane: 490.0 g
The percent yield of chloro-ethane in the reaction is 82.98%.
Answer:
Explanation:
2NO₂ ⇌ N₂O₄
E/mol·L⁻¹: 0.058 0.012
K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\
\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}