Respuesta :
I'm assuming you mean how many grams of sodium does 8.99 grams of sodium bromide have, because Bromine would not have any sodium in it.
Then, just direct calculate 22.34
[tex](22.34 \div 100) \times 8.99 = 2.008366[/tex]
Then, just direct calculate 22.34
[tex](22.34 \div 100) \times 8.99 = 2.008366[/tex]
Answer:
[tex]m_{Na}=2.01g[/tex]
Explanation:
Hello,
In this case, given the by mass percent of sodium in sodium bromide::
[tex]\% Na=\frac{m_{Na}}{m_{NaBr}}*100\%=22.34 \%[/tex]
Solving for the mass of sodium, considering the constant composition of sodium in the sodium bromide, we obtain:
[tex]m_{Na}=\frac{\% Na*m_{NaBr}}{100\%} =\frac{22.34\%*8.99g}{100\%} \\\\m_{Na}=2.01g[/tex]
Best regards.