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\begin{tabular}{|c|c|c|c|c|}
\hline
\multirow{5}{*}{\begin{tabular}{l}
Calculate the \\
ratio of sodium \\
to element "Z" \\
for compound B.
\end{tabular}} & Compound & \begin{tabular}{c}
Mass [tex]$Na$[/tex] \\
(grams)
\end{tabular} & \begin{tabular}{c}
Mass [tex]$Z$[/tex] \\
(grams)
\end{tabular} & Formula \\
\hline
& A & 12.00 & 2.44 & [tex]$Na_3 Z$[/tex] \\
\hline
& B & 3.45 & 0.700 & ? \\
\hline
& C & 14.00 & ? & [tex]$Na_3 Z$[/tex] \\
\hline
& D & 25.00 & 30.47 & ? \\
\hline
\end{tabular}

Calculate the ratio:

A. [tex]$4.93: 1$[/tex]
B. [tex]$1.22: 1$[/tex]
C. [tex]$2.42: 1$[/tex]
D. [tex]$0.203: 1$[/tex]

Sagot :

To solve for the ratio of sodium (Na) to element "Z" in compound B, follow these steps:

1. Identify the given masses:
- The mass of sodium (Na) in compound B is 3.45 grams.
- The mass of element Z in compound B is 0.700 grams.

2. Determine the ratio:
- The ratio of sodium (Na) to element Z is calculated by dividing the mass of sodium by the mass of element Z.

[tex]\[ \text{Ratio (Na to Z)} = \frac{\text{Mass of Na}}{\text{Mass of element Z}} \][/tex]

[tex]\[ \text{Ratio (Na to Z)} = \frac{3.45\text{ grams}}{0.700\text{ grams}} \][/tex]

3. Calculate the ratio:
[tex]\[ \frac{3.45}{0.700} \approx 4.93 \][/tex]

So, the ratio of sodium to element Z for compound B is approximately 4.93:1. This means that in compound B, for every 1 part of element Z, there are approximately 4.93 parts of sodium.