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VikaD [51]
2 years ago
9

In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so

lution is added from a buret to a colorless, acidified solution of H2O2(aq) in an Erlenmeyer flask. (Note: At the end point of the titration, the solution is a pale pink color.)
At a certain time during the titration, the rate of appearance of O2(g) was 1.0 x 10-3 mol/(L⋅s). What was the rate of disappearance of MnO4- at the same time. (please explain it)

Options
6.0 x 10-3 mol/(L⋅s)

A

4.0 x 10-3 mol/(L⋅s)

B

6.0 x 10-4 mol/(L⋅s)

C

4.0 x 10-4 mol/(L⋅s)
Chemistry
1 answer:
pickupchik [31]2 years ago
8 0

Answer:

C. 4x10⁻⁴ mol / (Ls)

Explanation:

Based in the reaction:

5 H₂O₂(aq) + 2 MnO₄⁻(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 O₂(g)

2 moles of MnO₄⁻ disappears while 5 moles of O₂ appears.

If 5 moles appears in a rate of 1.0x10⁻³mol /(Ls), 2 moles will disappear:

2 moles ₓ (1.0x10⁻³mol /(Ls) / 5 moles) = <em>4x10⁻⁴ mol / (Ls)</em>

Right answer is:

C. 4x10⁻⁴ mol / (Ls)

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Explanation:

The reaction equation will be as follows.

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Therefore, calculate the number of moles of sodium as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

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              = 195.65 mol \times 96500 C

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Relation between electrical energy and Q is as follows.

               E = Q \times V

Hence, putting the given values into the above formula and then calculate the value of electricity as follows.

              E = Q \times V

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                 = 9.4 \times 10^{7} J

As 1 J = 2.77 \times 10^{-7} kWh

Hence,      \frac{9.4 \times 10^{7}}{2.77 \times 10^{-7}} kWh

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Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.

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2 years ago
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What volume in<br><br> L<br><br> of a 0.724 M Nal solution contains 0.405 mol of Nal ?
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Answer:

0.5594\ \text{L}

Explanation:

Mol of NaI = 0.405 mol

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Molarity is given by

M=\dfrac{\text{mol}}{\text{Volume of solution in }NaI}\\\Rightarrow \text{Volume of solution in }NaI=\dfrac{0.405}{0.724}\\\Rightarrow \text{Volume of solution in }NaI=0.5594\ \text{L}

The required volume is 0.5594\ \text{L}.

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How many moles of N2 are essential for generating 0.08 moles of Li3N in the given reaction?6Li N2 2Li3N
jeka57 [31]
<span>08 moles Li3N * 1mole N2/2moles Li3N = 0.04 </span>
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A 1.50-liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25°C. Without changing th
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Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

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This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

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