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Moles of Li2CO3 = 1.53/73.891 = 0.0207 mole
Since HCl is in excess, amount of CO2 will depend on the limiting reagent which is Li2CO3.
∴Moles of CO2 = Moles of Li2CO3 = 0.0207.
Answer:
B
Explanation:
so that way they can hide away from the predators
Answer:
The answer to your question is V2 = 434.7 l
Explanation:
Data
Volume 1 = V1 = 240 l Volume 2 = ?
Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K
Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa
Process
1.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/t2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (300)(240)(293) / (479)(101.325)
3.- Simplification
V2 = 21096000 / 48534.675
4.- Result
V2 = 434.7 l
