Answer:
1. 156.818182 kg
2. 22.525 gallons
3. 437.4 yds
Explanation:
1. 345/2.2 (since 2.2 lbs equals 1 kg you would divide that by the number of pounds to find kg).
2. 85 x 1.06 = 90.1 (to find the number of qts)
90.1/4 = 22.525 (there are 4 quarts in a gallon so you would divide by 4)
3. 400 x 39.37 = 15,748 (multiply number of meters by the number of inches)
15,748/36 = 437.4 (divide by 36 since you are moving to a bigger unit)
Answer:
The correct answer is option b.
Explanation:

The ionic product of water : 
![K_w=[H^+][OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
A pure water has equal concentration of hydrogen ions and hydroxide ions, hence neutral.
![K_w=[H^+][H^+]=[H^+]^2](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BH%5E%2B%5D%3D%5BH%5E%2B%5D%5E2)
The ionic product of water at 283 K = 
![K_w=[H^+]^2](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5E2)
![0.29\times 10^{-14}=[H^+]^2](https://tex.z-dn.net/?f=0.29%5Ctimes%2010%5E%7B-14%7D%3D%5BH%5E%2B%5D%5E2)
![[H^+]=5.385\times 10^{-8} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5.385%5Ctimes%2010%5E%7B-8%7D%20M)
The pH of the water at 283 k;
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![=-\log[5.385\times 10^{-8} M]=7.26](https://tex.z-dn.net/?f=%3D-%5Clog%5B5.385%5Ctimes%2010%5E%7B-8%7D%20M%5D%3D7.26)
A pure water has equal concentration of hydrogen ions and hydroxide ions,So water will e neutral at this temperature also.
The correct answer is option b.
Answer:
C
Explanation:
The final product is an RNA molecule.
PKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5
![ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}](https://tex.z-dn.net/?f=ka%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BCH3CH2COO%5E-%5D%7D%7B%5BCH3CH2COOH%5D%7D%20)
![[CH3CH2COO^-] ](https://tex.z-dn.net/?f=%5BCH3CH2COO%5E-%5D%0A)
= 0.05
![[CH3CH2COOH]](https://tex.z-dn.net/?f=%5BCH3CH2COOH%5D)
= 0.10
Therefore 1.259x10^-5 =
![\frac{[H^+][0.05]}{[0.1]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5B0.05%5D%7D%7B%5B0.1%5D%7D%20)
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore
![[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%20%20%5Cfrac%7B%281.259%2A10%5E-5%29%280.1%29%7D%7B0.05%7D%20%20)
Therefore
![[H^+]= 2.513*10^-5](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%202.513%2A10%5E-5)
pH= -log [

] = -log(2.513*10^-5)= 4.59.
I think variation.... have a great day