7.5 is the answer. You have to move the decimal 2 places to the right.
An atom of element has there subatomic particles namely, proton, electron and neutron. Here, for a neutral atom, number of proton is equal to number of electron (this is not in the case of ions), this is equal to atomic number of an atom. In an atom, nucleus contains protons and neutrons which is responsible for mass of the atom and electrons move around nucleus in fixed orbits. Thus, atomic mass of an atom is equal to sum of number of protons and neutrons.
Option (b): Proton is the particle in nucleus of an atom, whose total number is equal to atomic number of that atom.
(4) Option (b): Atoms of same element have same atomic number because mass number can be different for different isotopes of atom. Since, atomic number is equal to number of protons, thus, number of protons are same for all atoms of the same element.
(5) Option (d): Isotopes are defined as atoms of same element with same atomic number but different mass number. Thus, correct option is (d) mass numbers.
Answer:
Both A and B will be unreactive!
Explanation:
10=2, 8
18=2, 8, 8
Answer:
the standard enthalpy of formation of this isomer of C₈H₁₈ (g) = -375 kj/mol
Explanation:
The given combustion reaction
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O ............................(1)
Heat of reaction or enthalpy of combustion = -5099.5 kj/mol
from equation (1)
ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation of reactants)
Or, - 5099.5 = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)].................................(2)
Given ΔH⁰f(CO₂) = - 393.5 kj/mol & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0
Using equation (2)
ΔH⁰f(C₈H₁₈) = -621 kj/mol
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.
