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 = 
Explanation:
In a voltaic cell, oxidation reaction occurs at anode whereas reduction reaction occurs at the cathode.
Hence, the half-cell reaction taking place at anode and cathode will be as follows.
At anode (Oxidation) :
...... (1)
At cathode (Reduction) : 
So, in order to balance the half cell reactions, we multiply reduction reaction by 3. Hence, reduction reaction equation will be as follows.
........ (2)
Therefore, overall reaction will be sum of equations as (1) + (2). Thus, net reaction equation is as follows.
Let us start with the total area of the lawn. Area= width x length, ie, 21 x 20 = 420 sq. ft. Snow flakes per square foot per minute = 1350 So Snow flakes for 420 sq.feet per minute = 420 x 1350 = 567000. Snow flakes for 1 hour = 567000 x 60 = 34020000 (60 minutes) Weight of 34020000 snow flakes = 34020000 x 1.60 = 54432000mg. To convert it into kilograms, divide this number by 1000000 (1 kilogram = 1000000 milligrams) Thus 54432000/1000000 = 54.432 kilograms or 54 kilograms and 432 grams.
Oxidation number of fe in the compound...,
Let the oxidation number of (Fe) be x
The oxidation number of oxygen(o) is (-2)
; 3x + 4(-2) = 0...where zero is the total charge on the compound(fe3o4)
; 3x = 8...then divide both sides by 3
Oxidation number of fe is( 2.67 )