Answer:
The number of molecules= 1.33 × 10∧22 molecules
percentage of mercury = 87%
Explanation:
Given data:
mass of dimethyl mercury = 5.10 g
number molecules of dimethyl mercury in 5.10 g = ?
percentage of mercury = ?
Solution:
First of all we will calculate the molar mass of dimethyl mercury.
molar mass of HgC2H6 = 1×200.6 + 2×12 + 6×1 = 230.6 g/mol
we know that,
230.6 g of HgC2H6 = 1 mol = 6.02 × 10∧23 molecules.
so
For the 5.10 g of sample:
5.10 g/230 g/mol = 0.022 moles of HgC2H6
now we will multiply these number of moles with Avogadro number to get number of molecules in 5.10 g of sample.
0.022 × 6.02 × 10∧23 molecules = 0.133 × 10∧23 molecules or
1.33 × 10∧22 molecules.
Percentage of mercury:
Formula:
percentage = (atomic number of Hg × total number of atoms of Hg/ molar mass of HgC2H6) × 100
% age of Hg = (200.6 g/mol× 1/ 230.6 g/mol) × 100
%age of Hg = 0.869 × 100
%age of Hg = 86.99 % or 87 %
The eagle is not a producer its a consumer
Explanation:
The chemical equation is as follows.
And, the given enthalpy is as follows.
; 
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.
x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
C = Carbon
CO = Carbon Monoxide
Co = Cobalt
O2 = Oxygen
So the answer is (3) CO
