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olasank [31]
3 years ago
9

Separate this redox reaction into its component half-reactions. 3o2 4co

Chemistry
1 answer:
swat323 years ago
3 0
The equation is:
3 O₂ + 4 Co → 2 Co₂O₃
Oxidation half reaction:
Co → Co³⁺ + 3 e
Reduction half reaction:
O₂ + 4 e → 2 O²⁻
To balance the equation number of electrons lost must be equal to number or electrons gained so we must multiply oxidation half time 4 and reduction half times 3
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Which statement is true according to the kinetic theory?
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Answer:

<em>Molecules of different gases with the same mass and temperature always have the same average kinetic energy - E.</em>

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Answer:

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Explanation:

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How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g

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2 years ago
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How many grams are in 44.8 liters of nitrogen gas, n2?<br> a) 56g<br> b) 27g<br> c) 36g<br> d) 112g
kkurt [141]

a) 56g

<h3>Calculation:</h3>

At STP,

22.4 L of N₂ = 1 mol

We have given 44.8 L of N₂, therefore,

44.8 L of N₂ = \frac{44.8}{22.4}

                    = 2 mol

We know that,

1 mol of N₂ = 28 g

Hence,

2 mol of N₂ = 28 × 2

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Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.

Learn more about calculation at STP here:

brainly.com/question/9509278

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