No, ammonium chloride in water is not a physical reaction. It is a chemical reaction. Hope this helps!
Answer:
3.01 L
Explanation:
V
1
: 2.75L
T
1
:
18
∘
C
V
2
: ?
T
2
:
45
∘
C
If you know your gas laws, you have to utilize a certain gas law called Charles' Law:
V
1
T
1
=
V
2
T
2
V
1
is the initial volume,
T
1
is initial temperature,
V
2
is final volume,
T
2
is final temperature.
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:
2.75
291
=
V
2
318
By cross multiplying, we have...
291
V
2
= 874.5
Dividing both sides by 291 to isolate
V
2
, we get...
V
2
= 3.005...
In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.
Therefore,
V
2
= 3.01 L
Answer:
A. 1.62
B. 0.513
C. 0.162.
Explanation:
pH = pKa + log [ HPO42-] / [H2PO4-]
Assuming, pKa is 7.21
Ka = 6.2 x 10^-8
pKa = 7.21
A.
7.0 = 7.21 + log [HPO42-] / [HPO4-]
- 0.21 = log [ HPO42-] / [H2PO4-]
[ HPO42-] / [ H2PO4-] = 10^-0.21
= 0.617
[H2PO4-] / [HPO42-] = 1 / 0.617
= 1.62
B.
7.5 = 7.21 + log [HPO42-] / [H2PO4-]
0.29 = log [HPO42-] / [H2PO4-]
[HPO42-] / [H2PO4-] = 10^0.29
= 1.95
[H2PO4-] / [HPO42-] = 1/1.95
= 0.513
8.0 = 7.21 + log [HPO42-] / [ H2PO4-]
0.79 = log [HPO42-] / [H2PO4-]
[HPO42-] / [H2PO4-] = 10^0.79
= 6.17
[H2PO4-] / [HPO42-] = 1/6.17
= 0.162
