State how many significant figures are proper in the results of the following calculations: (a)$\left(106.7\right)\left(98.2\right)\mathbf{/}\left(46.210\right)\left(1.01\right)$(b)${\left(18.7\right)}^{\mathbf{2}}$(c)$\left(1.60\times {10}^{-19}\right)\left(3712\right)$.

Uncertainty is the doubt that exists about the result of a measurement, which means how much the values deviate from each other. The precision of the measurement is assessed with significant figures. According to the rule, a significant figure in results must be equal to the quantity with the least number of significant figures.

Given Data

Consider the given data as below.

Calculation (a):$(106.7)(98.2)/(46.210)(1.01)$

Calculation (b):${(18.7)}^2$

The result 106.7 has 4 significant figures.

The result 98.2 has 3 significant figures.

The result 46.210 has 5 significant figures.

The result 1.01 has 3 significant figures.

The smallest number of significant figures is 3.

The result of 3 significant figures is,-

$(106.7)(98.2)/(46.210)(1.01)=225$

The given result is ${(18.7)}^2$.

Hence, the result ${(18.7)}^2$has 3 significant figure.

$$\begin{gathered} {(18.7)}^2=18.7\times 18.7 \\ =350 \end{gathered}$$

Hence, the smallest number of significant figures is 3.

The given result is $\left(1.60\times {10}^{-19}\right)\left(3712\right)$.

Remember when a number is written in scientific notation, only significant figures are the numerical part only. Therefore,-

The number$\left(1.60\times {10}^{-19}\right)$ has 3 significant figures.

And the number $\left(3712\right)$has 4 significant figures.

Here, in the result the smallest number of significant figures is 3. Therefore,-

$\left(1.60\times {10}^{-19}\right)\left(3712\right)=5.94\times {10}^{-16}$