Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
Answer:
100 cg/1g
Step-by-step explanation:
1 cg = 0.01 g Multiply by 100
100 cg = 1 g
(a) is <em>wrong</em>. The correct conversion factor is 1000 cm³/1 L.
(b) is <em>wrong</em>. The correct conversion factor is 1000 mL/1 L.
(c) is <em>wrong</em>. The correct conversion factor is 1 m/10 dm.
Answer:4 days will it take for the snail to get to the garden.
Explanation:
Speed of the snail = 12 feet per day:
Distance between the garden and snail = 48 feet
4 days will it take for the snail to get to the garden.
Answer:
The answer to your question is: C₁₈ H₂₇ N O₃
Explanation:
Data
Carbon = 70.79 g
Hydrogen = 8.91 g
Nitrogen = 4.58 g
Oxygen = 15.72 g
Process
AT C = 12 g
AT H = 1 g
AT N = 14 g
AT O = 16 g
Carbon
12 g ------------------------ 1 mol
70.79 g ------------------------- x
x = (70.79 x 1) / 12
x = 5.9 mol of C
Hydrogen
1 g ----------------------- 1 mol
8.91 g --------------------- x
x = (8.91 x 1) / 1
x = 8.91 mol of H
Nitrogen
14 g ---------------------- 1 mol
4.58 g ------------------- x
x = (4.58 x 1) / 14
x = 0.33 mol
Oxygen
16 g ------------------------ 1 mol
15.72 g -------------------- x
x = (15.72 x 1)/16
x = 0.98
Divide by the lowest number of moles
Carbon 5.9 / 0.33 = 17.9 ≈ 18
Hydrogen 8.91 / 0.33 = 27
Nitrogen 0.33 / 0.33 = 1
Oxygen 0.98 / 0.33 = 2.9 ≈ 3
C₁₈ H₂₇ N O₃
<h3>
Answer:</h3>
0.95 atm
<h3>
Explanation:</h3>
We are given;
Initial pressure, P1 = 1.0 atm
Initial temperature, T1 =298 K (25°C + 273)
Initial volume, V1 = 0.985 L
Final temperature, T2 = 295 K (22°C + 273)
Final volume, V2 = 1.030 L
We are required to find final air pressure;
Using the combined gas law;
To get, P2 ;
= 0.95 atm
Therefore, the air pressure at the top of the mountain is 0.95 atm
