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:
See explanation
Explanation:
The molecular geometry of an atom is connected to the number of electron pairs that surround it(whether lone pairs or bonding pairs) as well as its hybridization state. We shall now examine the N, P, or S atoms in each of the following compounds.
a)
In H3PO4, P has a tetrahedral molecular geometry and is sp3 hybridized.
b) In NH4NO3
N is sp3 hybridized in NH4^+ and sp2 hybridized in NO3^-. Also, N is tetrahedral in NH4^+ but trigonal planar in NO3^-.
c) In S2Cl2, we expect a tetrahedral geometry but as a result of the presence of two lone pairs on each sulphur atom, the molecular geometry is bent. The sulphur is sp3 hybridized.
d) In K4[O3POPO3], each phosphorus atom is in a tetrahedral molecular geometry and is sp3 hybridized.
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 $\boxed{\mathbf{3.6}}$}