Explanation:tr
a) Molar mass of HF = 20 g/mol
Atomic mass of hydrogen = 1 g/mol
Atomic mass of fluorine = 19 g/mol
Percentage of an element in a compound:

Percentage of fluorine:

Percentage of hydrogen:

b) Mass of hydrogen in 50 grams of HF sample.
Moles of HF = 
1 mole of HF has 1 mole of hydrogen atom.
Then 2.5 moles of HF will have:
of hydrogen atom.
Mass of 2.5 moles of hydrogen atom:
1 g/mol × 2.5 mol = 2.5 g
2.5 grams of hydrogen would be present in a 50 g sample of this compound.
c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.
Then mass of hydrogen in 50 grams of HF compound we will have :
5% of 50 grams of HF = 
Sodium , bromine zinc magnesium sulphur nitrogen potassium oxygen lead
Number of charge = 5018 C
<h3>Further explanation</h3>
Given
1000 C of charge for x grams of copper
Required
Number of charge
Solution
Faraday's Law :

For 1000 C, W = x grams

For 5x grams :

1 mole=6.02 ×10
8.3 moles=?
8.3×6.02×10=499.66 molecules
therefore ghe no. of molecules = 499.66
Answer:
correct option is (a)
The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over
(C6H5COOH)
Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)
Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350
X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5
x = V2.275 x 10^-5
X = 1.5083 x 10^-5 moles per liter H+
pH = -log(H+) = -log 1.5083 x 10^-5 which
= 4.6215