The nature of reactants, temperature, concentration, surface area and catalysts.
Yes, refer to the previous answer.
Answer:- C. H
Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.
As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.
The oxidation number in elemental form is zero.
In 
, the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in 
is -1. On product side, the oxidation number of hydrogen in 
is zero and in 
the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in 
is 0.
From above data, Oxidation number of O is -2 on both sides so it is not reduced.
Oxidation number of Cl is changing from -1 to 0 which is oxidation.
Oxidation number of H is changing from +1 to 0 which is reduction.
So, the right choice is C.H
Answer:
0.779 moles.
Explanation:
Mass of chlorine/number of moles = molar mass of chlorine.
Molar mass of chlorine = (35.5*2)
= 71 g/mol.
Mass = 55.3 g of chlorine.
Number of moles = 55.3/71
= 0.779 moles.
Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
