Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
A monobromination reaction of an alkane involves an alkane and bromine. The position of the hydrogen atom that will be substituted by the bromine free radical will depend on the order of the alkane. The bromine will attach to the carbon that has the most substituents.
a. pH=2.07
b. pH=3
c. pH=8
<h3>Further explanation</h3>
pH=-log [H⁺]
a) 0.1 M HF Ka = 7.2 x 10⁻⁴
HF= weak acid
![\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D%5Csqrt%7B7.2.10%5E%7B-4%7D%5Ctimes%200.1%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D8.5%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~8.5%3D2.07)
b) 1 x 10⁻³ M HNO₃
HNO₃ = strong acid
![\tt pH=-log[1\times 10^{-3}]=3](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-3%7D%5D%3D3)
c) 1 x 10⁻⁸ M HCl
![\tt pH=-log[1\times 10^{-8}]=8](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-8%7D%5D%3D8)
