Answer:
Complete combustion of 0.1mol of monocarboxylic acid requires V litres of O2 at dtc. and the reaction produces 0.3mol of CO2 and 0.2mol of H2O.
What is the value of V?
Explanation:
The balanced chemical equation of the reaction is
1mol. of acid reacts with 1mol of 
and produce 2mol. of CO2 and 2mol. of H2O.
Given,
0.1mol of acid requires how many litres of oxygen at STP?
0.1mol of acid requires 0.1mol of oxygen.
Since, 1mol of oxygen at STP occupies 22.4L of volume.
Hence, the volume occupied by 0.1mol of oxygen is -- 2.24L
Answer is:
The value of V is 2.24L.
<span>Answer: 0.00649M
The question is incomplete,
</span>
<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
</span>
<span>
With that you can solve the question following these steps"
</span>
<span>1) First ionization:
</span>
<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
</span>
<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>
<span>Do the mass balance:
</span>
<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />
<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
</span><span />
<span>Using the quadratic formula: x = 0.00649
</span><span />
<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Answer:
increased only 6 percent
Explanation:
The six pollutants regulated under the Clear Air Act increased only 6 percent between 1970 and 2001, while energy consumption increased 42 percent.
Answer:
7.22 L
Explanation:
From the question,
Applying Charles law,
V₁/T₁ = V₂/T₂...................... Equation 1
Where V₁ = Initial volume of gas, V₂ = Final volume of gas, T₁ = Initial Temperature of gas in Kelvin, T₂ = Final Temperature of gas in Kelvin
Make V₂ the subject of the equation
V₂ = (V₁×T₂)/T₁................. Equation 2
Given: V₁ = 5.00 L, T₁ = 20 °C = (20+273) K = 293 K, T₂ = 150 °C = (150+273) K = 423 K
Substitute these value into equation 2
V₂ = (5.00×423)/293
V₂ = 7.22 L
