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Sergeu [11.5K]
3 years ago
10

A rectangle has a width of 7 inches height of 14 inches and a length of 12 inches find it’s volume 

Chemistry
1 answer:
Nikolay [14]3 years ago
8 0
The answer would be 1,176. you solve by using (L*H*W)
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Por lo tanto, el compuesto contiene un átomo de carbono y dos átomos de azufre.

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melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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If your really kind please help I just want help
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