<u>We are given:</u>
M1 = 3 Molar V1 = 80 mL
M2 = x Molar V2 = 100 mL
<u>Finding the molarity:</u>
We know that:
M₁V₁ = M₂V₂
where V can be in any units
(3)(80) = (x)(100)
x = 240/100 [dividing both sides by 100]
x = 2.4 Molar
You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:
PH = Pka + log[acid/base]
Step 1 (Figure out the concentrations):
0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3
Step 2 (Plug into equation):
PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]
PH = 3.69
There would be 55.8 g present
Answer:
v = 534.5mL
m = 597.15g
Density = 9.23g/mL
Density = 9.125g/mL
Explanation:
Density = mass/ volume
For the first question
Density = 1.59g/mL
Mass = 834.01g
Volume = ?
Using the above formula we have 1.59 = 834.01/v
v = 834.01/1.59
v = 534.5mL
For the second question
Density =0.9167g/mL
Volume = 651.41mL
Mass =?
Using the above formula we have
0.9167 =m/651.41
Cross multiply
m = 0.9167 x 651.41
m = 597.15g
For the third question
Mass =803.44g
Volume=87.03mL
Density =?
Density = 803.44/87.03
= 9.23g/mL
For the fourth
Density = 56.85/6.23
= 9.125g/mL
Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
