Answer:
[Acetic acid] = 0.07 M
[Acetate] = 0.13 M
Explanation:
pH of buffer = 5
pKa of acetic acid = 4.76
![pH=p_{Ka} + log\frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3Dp_%7BKa%7D%20%2B%20log%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now using Henderson-Hasselbalch equation
![5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}](https://tex.z-dn.net/?f=5%3D4.76%20%2B%20log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D)
![log\frac{[Acetate]}{[Acetic\;acid]} = 0.24](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%200.24)
![\frac{[Acetate]}{[Acetic\;acid]} = 1.74](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%201.74)
....... (1)
It is given that,
[Acetate] + [Acetic acid] = 0.2 M ....... (2)
Now solving both the above equations
[Acetate] = 1.74[Acetic acid]
Substitute the concentration of acetate ion in equation (2)
1.74[Acetic acid] + [Acetic acid] = 0.2 M
[Acetic acid] = 0.2/2.74 = 0.07 M
[Acetate] = 0.2 - 0.07 = 0.13 M
I think that it is 1.5 mole it might not be
Answer:
A. 3.7 x 10⁻³
Explanation:
In the beginning of the reaction there are 0.200 moles of reactant. After 25 minutes, remain 0.108 moles. That means the moles that wer descomposed are:
0.200 moles - 0.108 moles = 0.092 moles of reactant were descomposed.
That descomposition occurs in 25 minutes. The average rate of descomposition in moles / minute are:
0.092 moles Methyl isonitrile / 25 minutes = 3.7x10⁻³ mol/min.
Right option is:
<h3>A. 3.7 x 10⁻³</h3>
Answer:
1). 1 mole of Carbon burnt in air
C + O2 →CO2
1 mole of carbon produces 1 mole of CO2 which is 44g of CO2
2). 1 mole of carbon is burnt in 16g of dioxygen
32g of O2 = 44g of CO2
1g of O2 = 44/32
CO2 (Dioxygen is limiting reagent)
16g of O2 = 4/32 × 16 = 22g of CO2 in one mole
3) 2 moles of Carbon burnt in 16g of dioxygen
16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2
