Zn⁰ ----> Zn⁺² + 2e⁻ - oxidation
Hg⁺² + 2e⁻ ----> Hg⁰ - reduction
Zn loses 2 moles of electrons , and Hg gains 2 mole of electrons.
So, number of moles of electrons gained and lost during reaction is equal.
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):

mass O₂ = 120 g
mol O₂(MW=32 g/mol) :

Mol ratio of reactants(to find limiting reatants) :

mol of H₂O based on O₂ as limiting reactants :
mol H₂O :

mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :

I think its c, it makes sense to me to be c
Answer:
Catalyst speeds up the chemical reactions. Catalyst is the compound that adds to the chemical reaction so that it happen more quickly. Catalyst without being consumed in the reaction accelerates the reaction. It lowers down the activation energy of the transition state and thus accelerates the reaction.
Hope this helped!
Plz mark me as Brainliest!
Combustion is a chemical reaction between a fuel and an oxidant, oxygen, to give off combustion products and heat. Complete combustion results when all of the fuel is consumed to form carbon dioxide and water, as in the case of a hydrocarbon fuel. Incomplete combustion results when insufficient oxygen reacts with the fuel, forming soot and carbon monoxide.
The complete combustion of propane proceeds through the following reaction:

+

-->

+

Combustion is an exothermic reaction, which means that it gives off heat as the reaction proceeds. For the complete combustion of propane, the heat of combustion is (-)2220 kJ/mole, where the minus sign indicates that the reaction is exothermic.
The molar mass of propane is 44.1 grams/mole. Using this value, the number of moles propane to be burned can be determined from the mass of propane given. Afterwards, this number of moles is multiplied by the heat of combustion to give the total heat produced from the reaction of the given mass of propane.
14.50 kg propane x <u> 1000 g </u> x <u> 1 mole propane </u> x <u> 2220 kJ </u>
1 kg 44.1 g 1 mole
=
729,931.97 kJ