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strojnjashka [21]
3 years ago
9

How many moles of electrons are required to reduce one mole of MnO4- to one mole of Mn2+? (2 points)

Chemistry
1 answer:
Dafna1 [17]3 years ago
8 0
The answer would be 6 e-. This is becuase you are turning a charge of -4 into a +2. In order to do this, you transfer 4 electrons for a neutral charge, and an additional 2 for a charge of +2.

This makes a total charge of +2, and the total transferred electrons 6 e-
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4 0
3 years ago
How many grams of lead (ll) sulfate would be produced from the complete reaction of 23.6 g lead (iV) oxide ?
djverab [1.8K]

Answer:

29.92grams of PbSO4

Explanation:

lead (iV) oxide = PbO2 = Molar mass: 239.2 g/mol

lead (ll) sulfate = PbSO4 = Molar mass: 303.26 g/mol

PbO2 = PbSO4

1:1 ratio

Pb = Lead

Lead has an oxidation number of 4+

O = Oxygen

Oxygen has an oxidation number of 2-

PbO2 + 4H+ + SO4 2- + 2e- = PbSO4(s) + 2H2O

Ok so the above would be the likely complete reaction, though we don't really need this as we already know the ratio is 1:1.

23.6g of PbO2

23.6/239.2 = 0.09866 Moles of PbO2

Since we have a 1:1 ratio we know that the same number of moles of PbSO4 are produced and since we know the molar mass it's simply molar mass multiplied by number of moles.

303.26 x 0.09866 = 29.92grams of PbSO4

6 0
3 years ago
What are the respective concentrations (m) of fe3+ and i- afforded by dissolving 0.200 mol fei3 in water and diluting to 725 ml?
Natalka [10]

Molarity = moles / liter of solution

Given, Moles of Fe³⁺ = 0.200

Volume of solution = 725 ml = 0.725 L

Conversion factor: 1000 ml = 1L

Molarity = 0.200 / 0.725 L = 0.275 M

The dissociation of Fel₃ in water is ad follows:

Fel₃ → Fe³⁺ + 3l⁻  

1 mole of Fel₃ gives 1 mole of Fe³⁺ ions  and 3 moles of l⁻  

Since the solution is 0.275 M of Fel₃, so there are 0.275 M of Fe³⁺ ions,

and (3 x 0.275 M) = 0.825 M of l⁻ ions.

7 0
3 years ago
A sample of gas occupies 10 L at STP. What
puteri [66]

Pressure is 5.7 atm

<u>Explanation:</u>

P1 = Standard pressure = 1 atm

P2 = ?  

V1 = Volume = 10L

V2= 2.4L

T1 = 0°C + 273 K = 273 K

T2 = 100°C + 273 K = 373 K

We have to find the pressure of the gas, by using the gas formula as,

$\frac{P 1 V 1}{T 1}=\frac{P 2 V 2}{T 2}

P2 can be found by rewriting the above expression as,

$P 2=\frac{P 1 \times V 1 \times T 2}{T 1 \times V 2}

Plugin the above values as,

$P 2=\frac{1 \text {atm} \times 10 L \times 373 \mathrm{K}}{273 \mathrm{K} \times 2.4 \mathrm{L}}=5.7 \text { atm }

4 0
3 years ago
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