<span>Answer: 0.00649M
The question is incomplete,
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<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
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<span>
With that you can solve the question following these steps"
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<span>1) First ionization:
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<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)
Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M
2) Second ionization
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<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
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<span>Do the mass balance:
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<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
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<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
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=> Ka₂ = (x²) / (0.01 - x) = 0.012
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<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
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<span>Using the quadratic formula: x = 0.00649
</span><span />
<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Answer:
so you can do the maths your self
Explanation:
isotopes are elements with he same proton number but difference in their neutron number
- proton number is also known as the atomic number
- mass number is the sum of the proton number and the neutron number
- mass number = proton number + neutron number
Answer:
1.2 liters.
Explanation:
Focus on the 4th digit: that's the ones column. The 3rd digit is the decimal place, just be sure to round up.
Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
Answer:
Explanation:
Chlorine is the element of group 17 and third period. The atomic number of chlorine is 17 and the symbol of the element is Cl.
The electronic configuration of the element chlorine is:-
Chloride ion is formed when chlorine atom gain one more electron. So, the ground-state electron configuration for the chloride ion is:-
