Answer:
The answer to your question is P2 = 0.78 atm
Explanation:
Data
Temperature 1 = T1 = 263°K Temperature 2 = T2 = 298°K
Volume 1 = V1 = 24 L Volume 2 = V2 = 35 L
Pressure 1 = P1 = 1 Pressure 2 = P2 = ?
Process
1.- To solve this problem use the Combined gas law
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (1)(24)(298) / (263)(35)
-Simplification
P2 = 7152 / 9205
-Result
P2 = 0.777
or P2 = 0.78 atm
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
Answer: 16.32 g of as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.34 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of as excess reagent are left.
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