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3 years ago

A 13.0 kg iron weightlifting plate has a volume of 1650 cm3 . What is the density of the iron plate in g/cm3?

Chemistry

2 answers:

Salsk061 [2.6K]3 years ago

8 0

11.0 kg = (11.0 kg)(1000 g/kg) = 11000 g
(11000 g)/(1400 cm3) = 7.857 g/cm3
Simplified = 7.86 g/cm3

Oksi-84 [34.3K]3 years ago

5 0

Explanation:

It is known that density is the amount of mass divided by volume.

Mathematically, Density = $\frac{mass}{volume}$

It is given that mass is 13.0 kg or 13000 g (as 1 kg = 1000 g). And, volume is 1650 $cm^{3}$ .

Therefore, calculate the density as follows.

Density = $\frac{mass}{volume}$

= $\frac{13000 g}{1650 cm^{3}}$

= $7.87 g/cm^{3}$

Thus, we can conclude that density of the given iron plate is $7.87 g/cm^{3}$ .

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1. Gather all the information in one place with molar masses above the formulas and masses below them.

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The molar ratio is 4 mol NO:4 mol NH₃

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From O₂:

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The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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