Answer: 3) 39.96 amu
Explanation:
Mass of isotope Ar- 36 = 35.97 amu
% abundance of isotope Ar- 36= 0.337% = 
Mass of isotope Ar- 38 = 37.96 amu
% abundance of isotope 2 = 0.063 % = 
Mass of isotope Ar- 40 = 39.96 amu
% abundance of isotope 2 = 99.600 % = 
Formula used for average atomic mass of an element :

![A=\sum[(35.97\times 3.37\times 10^{-3})+(37.96\times 6.3\times 10^{-4})+(39.96\times 0.996)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2835.97%5Ctimes%203.37%5Ctimes%2010%5E%7B-3%7D%29%2B%2837.96%5Ctimes%206.3%5Ctimes%2010%5E%7B-4%7D%29%2B%2839.96%5Ctimes%200.996%29%5D)

Therefore, the average atomic mass of argon is 39.96 amu
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
As Density = Mass/Volume
Mass = 26.0g
Density = 1.44g/mL
Therefore Volume = Mass/Density
=> Volume = 26.0/1.44 = 18.055... = 18.1mL (to 3 sig figs)
Answer:
H2S(g) + 2OH^-(aq) --------> S^2-(aq) + 2H2O(l)
Explanation:
We know that the net ionic equation shows the major reaction that occurs in the reaction system.
The molecular reaction equation is;
H2S(g) + 2NaOH(aq) ------> Na2S(aq) + 2H2O(l)
The complete ionic equation is;
H2S(g) + 2Na^+(aq) + 2OH^-(aq) --------> 2Na^+(aq) + S^2-(aq) + 2H2O(l)
Net ionic equation;
H2S(g) + 2OH^-(aq) --------> S^2-(aq) + 2H2O(l)
Answer:
1.7927 mL
Explanation:
The mass of solid taken = 4.75 g
This solid contains 21.6 wt%
, thus,
Mass of
=
= 1.026 g
Molar mass of
= 261.337 g/mol
The formula for the calculation of moles is shown below:
Thus,

Considering the reaction as:

1 moles of
react with 1 mole of 
Thus,
0.003926 mole of
react with 0.003926 mole of 
Moles of
= 0.003926 mole
Also, considering:

Molarity = 2.19 M
So,

Volume = 0.0017927 L
Also, 1 L = 1000 mL
<u>So, volume = 1.7927 mL</u>