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 %
DiPhosphorus pentoxide or pentaoxide
Answer:
Mass of ethanol obtained is 260.682 grams.
Volume of the ethanol obtained is 0.3304 L.
Explanation:

Moles of glucose = 
According to the reaction, 1 mol of gluocse gives 2 moles of ethanol.
Then 2.833 moles of glucose will give :

Mass of 5.667 moles of an ethanol :
5.667 mol × 46 g/mol = 260.682 g
Volume of ethanol = V
Density of the ethanol = 0.789 g/mL



Mass of ethanol obtained is 260.682 grams.
Volume of the ethanol obtained is 0.3304 L.
Most of the mass of an atom is located in the (nucleus), and most of the volume is taken up by the (electron clouds).
The ones in the parenthesis are your answers.
Hope this helps!
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.