Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
D. One chimp cleaning and grooming the hair of another chimp
[H+] in first brand:
4.5 = -log([H+])
[H+] = 10^(-4.5)
[H+] in second brand:
5 = -log[H+]
[H+] = 10^(-5)
Difference = 10^(-4.5) - 10^(-5)
= 2.2 x 10⁻⁵
The answer is A.
The correct answer is 1.1 moles
