Since this has to do with titration, the measured volume of NaOH used would be less than the actual volume used because some of the unused NaOH would cling to the sides of the buret and therefore wouldn't factor into the measurement
Hope this helps
Answer:
ΔH = -110.5kJ
Explanation:
It is possible to obtain enthalpy of combustion of a particular reaction by the algebraic sum of similar reactions (Hess's law). Using:
1. C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5kJ
2. CO(g) + 1/2O₂(g) → CO₂(g) ΔH₂ = -283.0kJ
The sum of 1 -2 gives:
C(s) + <u>O₂(g)</u> + <u>CO₂(g)</u> → <u>CO₂(g)</u> + CO(g) + <u>1/2O₂(g)</u>
C(s) + 1/2O₂(g) → CO(g) ΔH = -393.5kJ - (-283.0kJ) =
<h3>ΔH = -110.5kJ</h3>
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Answer:
M = 0.441 M
Explanation:
In this case, we have two solutions that involves the Manganese II cation;
We have Mn(CH₃COOH)₂ and MnSO₄
In both cases, the moles of Mn are the same in reaction as we can see here:
Mn(CH₃COO)₂ <-------> Mn²⁺ + 2CH₃COO⁻
MnSO₄ <------> Mn²⁺ + SO₄²⁻
Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:
moles of MnAce = 0.489 * 0.0283 = 0.0138 moles
moles MnSulf = 0.339 * 0.0125 = 0.0042 moles
the total moles are:
moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles
Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L
M = 0.018 / 0.0408
M = 0.441 M
This would be the final concentration of the manganese after the mixing of the two solutions
In 1869 Russian chemist DIMITRI MENDELEEV started the development of the periodic table,arranging chemical elements by atomic mass. He predicted the discovery of other elements and left spaces open in his periodic table for them. HOPE THIS HELPSS HAVE A GREAT DAY <333
