The average atomic mass of Sn is 118.71 g/mol
the percentage of heaviest Sn is 5.80%
the given mass of Sn is 82g
The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691
Total atoms of Sn in 82g = 
the percentage of heaviest Sn is 5.80%
So the total atoms of
= 5.80% X 
Total atoms of
=
atoms
the mass of
will be = 
To solve this we use the
equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the
volume of the stock solution, M2 is the concentration of the new solution and
V2 is its volume.
14 M x V1 = 4.20 M x 200 mL
V1 = 60 mL needed of the concentrated solution
<u>Answer:</u> The equilibrium constant for
equation is 
<u>Explanation:</u>
The given chemical equation follows:

The value of equilibrium constant for the above equation is 
Calculating the equilibrium constant for the given equation:

The value of equilibrium constant for the above equation will be:

Hence, the equilibrium constant for
equation is 
Answer : The mole fraction of nitrogen will be 0.4615.
Explanation : When nitrogen (
)and hydrogen (
)are mixed, the mole ratio becomes 1 : 1.5,
Now we know that (
) is acting as a limiting agent.
So at the time of when 0.4 moles of (
) is been formed it requires 0.4 moles of (
) and 3.4 moles of (
)
So, we find the the remaining (
) will be 0.6 and
(
) will be 0.3 mole present in mixture.
So, the mole fraction of (
) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615
PH is calculated using <span>Handerson- Hasselbalch equation,
pH = pKa + log [conjugate base] / [acid]
Conjugate Base = Acetate (CH</span>₃COO⁻)
Acid = Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]
We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,
pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵
pKa = 4.75
Now,
Putting all values in eq. 1,
pH = 4.75 + log [0.172] / [0.818]
pH = 4.072