Question

Taking logarithms and antilogarithms is necessary to solve many chemistry problems. For practice, complete the following table, where is a number.

Answer 1

Given the question on the question tab;

`\begin{gathered} N=64.5 \\ LogN=log64.5=1.80955 \end{gathered}`
`\begin{gathered} N=? \\ LogN=-0.006 \\ \therefore N=antilog(-0.006) \\ N=10^(-0.006) \\ N=0.9863 \end{gathered}`
`\begin{gathered} N=5.40 \\ LogN=log5.40=0.73239 \end{gathered}`

Final answer: