Answer:
3
Explanation:
In scientific notation, the number of integers in the first part will always be the amount of significant figures.
For example,
1) 1.542505 x 10^17 =7 significant figures
1) 1.54 x 10^17 =3 significant figures
1. pesticides: that are sprayed on crops and flowers
2. global warming: warm winters have caused plants to shift their schedules. when the honey bees come out of hibernation, the flowers they need to feed on have already bloomed and died
3. pathogens carried by mites weaken bees
Answer: A pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.
Explanation:
The horizontal rows in a period table are called periods.
Elements present in the same period will have same atomic orbitals.
For example, electronic distribution of Na is 2, 8, 1 and it is a third period element.
Similarly, electronic distribution of Cl is 2, 8, 7 and it is also a third period element.
Hence, both Na and Cl will have K, L, M shells, that is, they have three atomic orbitals.
Thus, we can conclude that a pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.
The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H₂SO₄ contains more HSO₄⁻ ions than SO₄²⁻ ions is <u>True.</u> This is best explained by the fact that H₂SO₄ <u>is a diprotic acid where only the first hydrogen completely ionizes.</u>
Why?
H₂SO₄ is a diprotic acid. That means that it has <u>two hydrogen ions</u> to give to the solution. The two dissociation reactions are shown below:
H₂SO₄ + H₂O → HSO₄⁻ + H₃O⁺
HSO₄⁻ + H₂O ⇄ SO₄²⁻ + H₃O⁺
As the arrows show, the first dissociation is complete, meaning that all the sulfuric acid that is present initially is dissociated into HSO₄⁻ and H₃O⁺. However, the second dissociation is incomplete, and it's actually an equilibrium with an acid constant (Ka)of 1.2×10⁻².
That means that if the initial concentration of H₂SO₄ was 1M, the concentration of HSO₄⁻ is going to be 1M as well, but <u>the concentration of SO₄²⁻ is going to be much less than 1M</u>, according to the dissociation constant.
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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.
