Answer:
What do you mean Every force has a what?
Answer:
a) 10.457.
b) 9.32.
c) 8.04.
d) 6.58.
e) 4.76.
f) 2.87.
Explanation:
- Aziridine is an organic compounds containing the aziridine functional group, a three-membered heterocycle with one amine group (-NH-) and two methylene bridges (-CH2-). The parent compound is aziridine (or ethylene imine), with molecular formula C2H5N.
- Aziridine has a basic character.
- So, pKb = 14 – 8.04 = 5.96
- If we denote Aziridine the symbol (Az), It is dissociated in water as:
Az + H₂O → AzH⁺ + OH⁻
<u><em>a) 0.00 ml of HNO₃:
</em></u>
There is only Az,
[OH⁻] = √(Kb.C)
Kb = 1.1 x 10⁻⁶. & C = 0.0750 M.
[OH⁻] = √(1.1 x 10⁻⁶)(0.075) = 2.867 x 10⁻⁴.
∵ pOH = - log[OH-] = - log (2.867 x 10⁻⁴) = 3.542.
∴ pH = 14 – pOH = 14 – 3.542 = 10.457.
<u><em>b) 5.27 ml of HNO₃</em></u>
- To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
- No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
- No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (5.27 ml) = 0.302 mmol.
- The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (0.302).
- This will form a basic buffer in the presence of weak base (Az).
<em>pOH = pKb + log[salt]/[base]
</em>
- [salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (0.302) / (85.27) = 3.54 x 10⁻³ M.
- [base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 0.302 mmol) / (85.27 ml) = 0.0668 M.
- pOH = pKb + log[salt]/[base] = 5.96 + log[3.54 x 10⁻³]/[ 0.0668 M] = 4.68.
- <em>pH = 14 – pOH = 14 – 4.68 = 9.32.
</em>
<em />
<em><u>c) Volume of HNO₃ equal to half the equivalence point volume
:</u></em>
- At half equivalence point, the concentration of the salt formed is equal to the concentration of the remaining base (aziridine), [salt] = [base].
- pOH = pKb + log[salt]/[base] = 5.96 + log[1.0] = 5.96.
- pH = 14 – pOH = 14 – 5.96 = 8.04.
<u><em>d) 101 ml of HNO₃:
</em></u>
- To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).
- No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.
- No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (101.0 ml) = 5.7974 mmol.
- The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (5.7974).
- This will form a basic buffer in the presence of weak base (Az).
<em>pOH = pKb + log[salt]/[base]
</em>
- [salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (5.7974) / (181.0) = 0.032 M.
- [base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 5.7974 mmol) / (181.0 ml) = 0.00112 M.
- pOH = pKb + log[salt]/[base] = 5.96 + log[0.032]/[ 0.00112] = 7.416.
- pH = 14 – pOH = 14 – 7.416 = 6.58.
<u><em>e) Volume of HNO₃ equal to the equivalence point
:</em></u>
- At the equivalence point the no. of millimoles of the base is equal to that of the acid.
- Volume of HNO₃ needed for the equivalence point = (6.00 mmol) / (0.0574 mmol/ml) = 104.5 ml
At the equivalence point:
- [AzH⁺] = (6.00 mmol) / (80.0 + 104.5 ml) = 0.0325 M.
- As Ka is very small, the dissociation of AzH⁺ can be negligible.
Hence, [AzH⁺] at eqm ≈ 0.0325 M.
- [H+] = √(Ka.C) = √(10⁻⁸˙⁰⁴ x 0.0325) = 1.72 x 10⁻⁵.
- pH = - log[H+] = - log(1.72 x 10⁻⁵) = 4.76.
<u><em>f) 109 ml of HNO₃:
</em></u>
- No. of milli-moles of H⁺ added from HNO₃ = (0.0574 mmol/ml) × (109 ml) = 6.257 mmol.
- Which is higher than the no. of millimoles of the base (Az) = 6.0 mmol.
- After the addition, [H⁺] = (6.257 - 6.00) / (80.0 + 109 mL) = 0.00136 M.
- As Ka is small and due to the common ion effect in the presence of H⁺, the dissociation of Az is negligible.
- pH = -log[H⁺] = -log(0.00136) = 2.87.
produced from the reaction.
The overall balanced equation for the reaction is,
So, the number of molecules produced in the reaction of
produced from the reaction.
reaction.
are produced in the reaction.
To know more about moles, refer to the below link:
brainly.com/question/26416088
#SPJ4
Answer:
2Cs+Sr(CrO4)-->Cs2(CrO4)+Sr
Explanation:
just set them equal, only value that changed was the amount of cesium(Cs)
<span>R=K[NO]²[Cl2]</span><span>¹
When we look at the rate of reaction formula,
we can see exponent of [NO] =2, that means that order of reaction with respect to NO equals 2.
We can see that exponent of [Cl2] =1, that means that order of reaction with respect to Cl2 equals 1.
Overall order of reaction = 2+1 = 3
If concentration of NO is tripled, then
</span><span>R=K[NO]²[Cl2]
</span>R1=K(3[NO])²[Cl2]=9K[NO]²[Cl2]=9R.
If the concentration of NO is tripled, then rate will increase 9 times.