Molarity=moles/liters. Take 42.9 g CaCl2 and divide it by the molar mass of CaCl2 which is 110.98. You get .387 moles. Finally divide .387 moles/ 45 L = .0086 M CaCl2
The particles will move faster and further apart. There will be more energy in the particles due to the heat.
Its the 2nd one
Because a HYDROCARBON contains <u><em>only hydrogen and carbon</em></u>
<u><em>THE REST HAVE OTHERS IN THEM</em></u>
<u><em></em></u>
<u><em>HAVE A GREAT DAY :)</em></u>
Explanation:
The given data is as follows.
= 0.042 M,
for ![HP^{-} = 3.9 \times 10^{-6}](https://tex.z-dn.net/?f=HP%5E%7B-%7D%20%3D%203.9%20%5Ctimes%2010%5E%7B-6%7D)
According to the given situation
acts as a base.The reaction equation will be as follows.
![P^{2-} + H_{2}O \rightleftharpoons HP^{-} + OH^{-}](https://tex.z-dn.net/?f=P%5E%7B2-%7D%20%2B%20H_%7B2%7DO%20%5Crightleftharpoons%20HP%5E%7B-%7D%20%2B%20OH%5E%7B-%7D)
Relation between
and
are as follows.
![K_{a} \times K_{b} = K_{w}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%5Ctimes%20K_%7Bb%7D%20%3D%20K_%7Bw%7D)
![K_{b} = \frac{1 \times 10^{-14}}{K_{a}}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B1%20%5Ctimes%2010%5E%7B-14%7D%7D%7BK_%7Ba%7D%7D)
= ![\frac{1 \times 10^{-14}}{3.9 \times 10^{-6}}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%20%5Ctimes%2010%5E%7B-14%7D%7D%7B3.9%20%5Ctimes%2010%5E%7B-6%7D%7D%7D)
= ![2.6 \times 10^{-9}](https://tex.z-dn.net/?f=2.6%20%5Ctimes%2010%5E%7B-9%7D)
Also,
Let us take
= x
So, ![K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHP%5E%7B-%7D%5D%5BOH%5E%7B-%7D%5D%7D%7BP%5E%7B2-%7D%7D)
x = ![1.04 \times 10^{-5}](https://tex.z-dn.net/?f=1.04%20%5Ctimes%2010%5E%7B-5%7D)
= ![1.04 \times 10^{-5}](https://tex.z-dn.net/?f=1.04%20%5Ctimes%2010%5E%7B-5%7D)
pOH = - log![[OH^{-}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D)
= - log (
)
= 4.99
As it is known that pH + pOH = 14
so, pH + 4.99 = 14
pH = 9.01
Thus, we can conclude that pH of the solution is 9.01.
The new volume of the gas is 144 mL.
<u>Explanation:</u>
At STP, the temperature is 273 K an pressure is 1 atm.
1 atm=101kPa
Given that initial volume V1=36 mL
Initial pressure P1=1atm=101 kPa
Final pressure P2=25.3kPa
We have to determine the final volume V2
According to Boyle's law
![P_{1} V_{1} =P_{2} V_{2}](https://tex.z-dn.net/?f=P_%7B1%7D%20V_%7B1%7D%20%3DP_%7B2%7D%20V_%7B2%7D)
![V_{2} =P_{1} V_{1} /P_{2}](https://tex.z-dn.net/?f=V_%7B2%7D%20%3DP_%7B1%7D%20V_%7B1%7D%20%2FP_%7B2%7D)
![=25.3\times101/25.3](https://tex.z-dn.net/?f=%3D25.3%5Ctimes101%2F25.3)
=144 mL
Final volume =144mL