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Nataly_w [17]
3 years ago
8

How would you prepare 3.5 L of a 0.9M solution of KCl?

Chemistry
1 answer:
PolarNik [594]3 years ago
4 0
V=3,5L\\
Cm=0,9M\\
M_{KCl}=74\frac{g}{mol}\\\\
C_{m}=\frac{n}{V}\\\\
n=\frac{m}{M}\\\\
C_{m}=\frac{m}{MV} \ \ \ \Rightarrow  \ \ \ m=C_{m}MV\\\\
m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark. 
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No' of molecules divide by avogadro number , 6×6.023×10^23 so (2.2×10^22)÷(6.023×10^23)
= 0.03653 moles
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n×Mr=m
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Answer:

2KBr + MgF₂ –> 2KF + MgBr₂

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KBr + MgF₂ –> KF + MgBr₂

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There are 2 atoms of F on the left side and 1 atom on the right. It can be balance by writing 2 before KF as shown below:

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Answer:

ver explicacion

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