Given:
P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?
Assume that the gas ideally, thus, we can use the ideal gas equation:
PV = nRT
where R = 0.0821 L atm/mol K
123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T
solve for T
T = 295.72 K<span />
Explanation:
a. LiCl is an ionic molecule whereas water is a polar molecule with net dipole moment in it. There LiCl in water would have an ion-dipole force of interaction.
b. Both NF3 and CH3CN have dipole moment in them, since both are polar molecule. Hence, there would be dipole-dipole interaction.
c. Here both CCl4 and benzene are non polar molecules therefore, they have London dispersion force of interaction.
d. In methylamine and water both have hydrogen bonding in them. The nitrogen of CH3NH2 forms hydrogen bond with water.
Answer : the hydrogen ion concentrations for pH = 7.35 and pH = 7.45 are 4.46 x 10⁻⁸ M and 3.54 x 10⁻⁸ M respectively.
When the pH of a solution is less than 7, the hydrogen ion concentration is higher and therefore the solution is acidic.
On the other hand, when the pH of a solution is greater than 7, hydroxide ion concentration is higher and the solution acts as a base.
The normal pH range of blood is given as 7.35 - 7.45 which is greater than 7. That means blood is slight basic in nature.
Let us find hydroxide ion concentration of blood.



![pOH = - log [OH-]](https://tex.z-dn.net/?f=pOH%20%3D%20-%20log%20%5BOH-%5D)
![[OH-] = 10^{-6.65}](https://tex.z-dn.net/?f=%5BOH-%5D%20%3D%2010%5E%7B-6.65%7D)
![[OH-] = 2.24 \times 10^{-7}](https://tex.z-dn.net/?f=%5BOH-%5D%20%3D%202.24%20%5Ctimes%2010%5E%7B-7%7D)
Ionic product of water is written as,
![[H+] [OH-] = 1 \times 10^{-14}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%5BOH-%5D%20%3D%201%20%5Ctimes%2010%5E%7B-14%7D)
Let us plug in the calculated value of [OH-]
![[H+] = \frac{1 \times 10^{-14}}{2.24 \times 10^-7} = 4.46 \times 10^{-8} M](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%20%5Cfrac%7B1%20%5Ctimes%2010%5E%7B-14%7D%7D%7B2.24%20%5Ctimes%2010%5E-7%7D%20%3D%204.46%20%5Ctimes%2010%5E%7B-8%7D%20M)
The hydrogen ion concentration for pH = 7.35 is 4.46 x 10⁻⁸ M
Calculation of [H+] for pH = 7.45
pOH = 6.55
[OH-] = 2.82 x 10⁻⁷
![[H+] = \frac{1 \times 10^{-14}}{2.82 \times 10^-7} = 3.54 \times 10^{-8} M](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%20%5Cfrac%7B1%20%5Ctimes%2010%5E%7B-14%7D%7D%7B2.82%20%5Ctimes%2010%5E-7%7D%20%3D%203.54%20%5Ctimes%2010%5E%7B-8%7D%20M)
Similarly, for pH = 7.45, we get hydrogen ion concentration of 3.54 x 10⁻⁸ M.
<span>The answer is C.) Layer A is the stratosphere and Layer B is the troposphere.
Let me know if you have any other questions ♥
</span>
The correct answer is C, too deep in the Earth to collect. Hope this helps!