Answer:
449.5 g
Explanation:
Silver sulfate- Ag2SO4
M(Ag)=107 g/mol => M(Ag2)=214 g/mol
M(S)=32 g/mol
M(O)=16 g/mol => M(O4)=64 g/mol
M(Ag2SO4)=310 g/mol
n=1.45 mol
m(Ag2SO4)=M(Ag2SO4)*n=310 g/mol *1.45 mol= 449.5 g
The answer is G Container 2
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Hope this helps
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Zane
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<span> esskeetit</span>
To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.
A reaction rate is generally measured in terms of amount (usually in mass) per unit volume.