Answer:
A) pH of the buffer is 3.44
B) pH of the buffer solution is 3.37
Explanation:
Relation between K_{a} and pK_{a} is as follows.
.
The relation between pH and pK_{a} is as follows.
![pH = pK_{a} + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
pH of the buffer is 3.44.
b)
mol of HCl added = 11.6M *0.001 L = 0.0116 mol
In the given reaction, 
will react with 
to form
No. of moles of
And, no. of moles of 
= 0.12 mol
after the reaction :
No. of moles of = moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of = moles present initially + moles added
= (0.12 + 0.0116)
= 0.1316 mol
As,
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
![pH = pK_{a} + log \frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
= 3.347+ log {0.1384/0.1316}
= 3.369
≅ 3.37
pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37
Answer:
For 1 mole of Lead(II)chlorie we have 1 mole of lead (Pb), 2 moles of Chlorine (Cl) and 4 moles of oxygen (O).
Explanation:
<u>Step 1: </u>Data given
Number of moles of Pb(ClO2)2 = 1
Since the oxidation number of Pb is 2+ and the oxidation number of ClO2 is -1
So to bind we should have <u>twice</u> ClO2
1 mole of Pb(ClO2)2 contains 1 mole of Pb; 1*<u>2</u>= 2 moles of Cl and 2*<u>2</u> = 4 moles of 0.
This means for 1 mole of Lead(II)chlorie we have 1 mole of lead (Pb), 2 moles of Chlorine (Cl) and 4 moles of oxygen (O).
Answer:
Option B. 3 m/s east
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 2 m/s east
Final velocity (v) = 5 m/s east
Time (t) = 6 s
Change in velocity =?
We can obtain the change in Lee's velocity as illustrated below:
Initial velocity (u) = 2 m/s east
Final velocity (v) = 5 m/s east
Change in velocity =?
Change in velocity = Final velocity (v) – Initial velocity (u)
Change in velocity = v – u
Change in velocity = 5 – 2
Change in velocity = 3 m/s east
Thus, the change in Lee's velocity is 3 m/s east
Answer:
8kJ/mol
Explanation:
since the forward reaction is -8kJ/mol, the backward reaction has the same enthalphy but reversed
The answer is C
6.0m x 8.0m= 48.0m
then converting it to cm by multiply
48 x 100
