Question

What is the coefficient?
A. A whole number that appears in front of a formula in a balanced chemical equation.
B. A whole number that appears as a subscript at the end of a formula in a balanced chemical equation.
C. A whole number that appears as a subscript in front of a formula in a balanced chemical equation.
D. A fraction that appears in front of a formula in a balanced chemical equation.
E. A whole number that appears as a superscript at the end of a formula in a balanced chemical equation

Answer 1

To answer this question, you should recall the concept of a coefficient. A coefficient is an expression denoting how many molecules or moles of a given formula are present in a chemical reaction.

Complete step by step answer:
The coefficient can be defined as a small whole number that appears as a factor in front of a formula in a chemical equation. It indicates relative amounts of reactants and products. It is also evidence of the number of moles of reactant and product. Coefficient multiples the number of atoms of each element in the formula and it must be the smallest ratio possible.

Hence, the correct answer to this question of option A.

Note:
In Algebra the Coefficient is the number in front of the variable terms. In any type of chemical equation, there are two types of numbers. They are subscripts and coefficients. Subscripts are defined as a part of the chemical formulas of the reactants and products and the other type talked above are coefficients that are placed in front of the formulas to indicate how many molecules of that substance is used or produced. While the other type coefficients indicate the number of each substance involved in the reaction and may be changed to balance the equation. In the Haber’s process equation for Ammonia ${N_2} + 3{H_2} \to 2N{H_3}$. Here we can see that coefficients of:
1 for ${N_2}$, 3 for ${H_2}$and 2 for $N{H_3}$. The reaction shows that each cycle of the forward reaction involves 1 molecule of ${N_2}$ with 3 molecules ${H_2}$ to produce 2 molecules of $N{H_3}$. Subscripts in the reaction cannot be changed.