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Alexxandr [17]
3 years ago
9

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

cket motors. The reaction is the following. 3 NiO(s) + 2 Al(s) → 3 Ni(l) + Al2O3(s) What masses of nickel(II) oxide and aluminum must be used to produce 14.8 g of nickel
Chemistry
1 answer:
melamori03 [73]3 years ago
4 0

<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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Answer:

The Aluminium sample

Explanation:

From the question,

ΔQ of the iron = Cm(t₂-t₁)........................ Equation 1

Given: C = specific heat capacity of iron, m = mass of iron, t₁ and t₂ = initial and final temperature respectively.

Given: m = 175 g = 0.175 kg, t₂ = 21.5°C, t₁ = 99.7°C

Constant: C = 444 J/kgK.

Substitute into equation 1

ΔQ = 0.175(444)(21.5-99.7)

ΔQ = -6076.14 J

Similarly, for aluminium,

ΔQ' = c'm'(t₂-t₁)...................... Equation 2

Given: m' = 175 g = 0.175 kg,

Constant: 900 J/kgK

ΔQ' = 0.175(900)(21.5-99.7)

ΔQ' = -12316.5 J

Hence the aluminium sample undergoes the greater heat change

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3 years ago
What is %N in (NH4)3PO4
Salsk061 [2.6K]
<h3>Answer:</h3>

28.17 %

<h3>Explanation:</h3>

In the question we are required to determine the percent by mass of nitrogen in  (NH₄)₃PO₄.

  • To calculate the percent by mass of an element in a compound we use the formula;
  • % by mass =(Mass of the element÷ Relative formula mass of the compound)100

In this case;

  • Atomic mass of nitrogen is 14.0 g/mol

But; There are three nitrogen atoms in (NH₄)₃PO₄.

Therefore; Mass of nitrogen in (NH₄)₃PO₄ = 14.0 × 3

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The relative formula mass of (NH₄)₃PO₄ = 149.0867 g/mol

Thus;

% by mass of N = (42 g ÷ 149.0867 g/mo)× 100%

                          = 28.17 %

Therefore, the % by mass of N is (NH₄)₃PO₄ is 28.17%

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Butter has a specific gravity of 0.86. What is the mass of 2.45 L<br> of butter?
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Answer:

mass of the butter is 2.107 kg

Explanation:

specific gravity = destiny of the substance / density of

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0.86 = density of the butter / 1000

where density of the water is 1000 kg/m^3 at 4°C

density of the butter = 860kg/m^3

now,

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860 = mass of the butter / 2.45 × 10^-3

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. Citric acid, which can be obtained from lemon juice, has the molecular formula C6H8O7. A 0.250-g sample of citric acid dissolv
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n = 0.250/192

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The dissociation reaction of one molecule of  the acid will give the stoichiometry:

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0.0013 mol --------------------------- 0.0039

For a simple direct three rule:

0.0013x = 0.0039

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