A basketball is inflated to a pressure of 1.50 atm in a 20.0°C garage. What is the pressure of the basketball outside
2 answers:
Answer: P = 1.37 atm
Explanation: For this problem we will use the Gay Lussac Law which is P1/T1 = P2/T2 then derive for P2 which is the outside pressure. Also remember to convert the units of temperature from °C to Kelvin.
T1 = 20°C + 273 = 293 K
T2 = -5°C + 273 = 268 K
P2 = P1 T2 / T1
= 1.50 atm ( 268 K ) / 293 K
= 1.37 atm
Cancel out the units of K so that the remaining unit will be in atm.
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Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.
Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
The half-life of the reaction is 50 minutes
Data;
- Time = 43 minutes
- Type of reaction = first order
- Amount of Completion = 45%
Let the initial concentration of the reaction be X
The reactant left = (1 - 0.45) X = 0.55 X
= X
For a first order reaction
The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.
Substitute the values
The half-life of the reaction is 50 minutes
Learn more on half-life of a first order reaction here;