Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess
P4)
<h3>How to calculate moles?</h3>
The balanced chemical equation is

The mass of clorine is m(
) = 58.0 g
The amount of clorine is n(
) = m(
)/M(
) = 58/70.906 = 0.817 mol
The stoichiometric reaction,shows that
10 moles of
yield 4 moles of
;
0.817 of
yield x moles of 
n(
) = 4*0.817/10 = 0.3268 mol
To know more about stoichiometric reaction, refer:
brainly.com/question/14935523
#SPJ9
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Given:
Putting in the values we get:
Thus the rate of appearance of
is 0.151
Answer: Partial pressure of
at a depth of 132 ft below sea level is 2964 mm Hg.
Explanation:
It is known that 1 atm = 760 mm Hg.
Also, 
where,
= partial pressure of 
P = atmospheric pressure
= mole fraction of 
Putting the given values into the above formula as follows.


= 0.780
Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of
is as follows.

= 
= 2964 mm Hg
Therefore, we can conclude that partial pressure of
at a depth of 132 ft below sea level is 2964 mm Hg.