How many grams of calcium bromide are in 50.0 mL of a 0.25 M calcium bromide solution?

1 answer:

8 0

$C_{m}=\frac{n}{V_{r}}\\\\
m=\frac{m}{M}\\\\
C_{m}=\frac{m}{MV_{r}} \ \ \ \ \Rightarrow \ \ \ \ m=C_{m}MV_{r}}\\\\
C_{m}=0.25\frac{mol}{L}\\
M_{CaBr_{2}}=200\frac{g}{mol}\\
V_{r}=50mL=0.05L\\\\
m=0.25\frac{mol}{L}*200\frac{g}{mol}*0.05L= \emph{\textbf{2.5g}}$

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Ea2=89.28JKoule/mol

<h3>What is the activation energy of the new pathway, all other factors being equal?</h3>

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