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lesya [120]
3 years ago
15

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

bers plugged in with proper UNITS, and clearly indicate your final answer).
Mass=1.3x1022kg, Diameter=2300km

B) (extra credit problem) Working backwards... Calculate the mean radius of the following object (if it was a perfect sphere).

Density=687kg/m3, Mass=5.68x1026kg
Chemistry
1 answer:
-BARSIC- [3]3 years ago
3 0

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}

V = \frac{ 4}{3 }\pi r^{3}

r^{3} = \frac{3V }{4 \pi }\

r= \sqrt [3]{ \frac{3V }{4 \pi } }

r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}

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How many grams of Fe can be produced when 5.50 g of Fe2O3 reacts?
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Answer:

3,85 g of Fe

Explanation:

1- The first thing to do is calculate the molar mass of the Fe2O3 compound. With the help of a periodic table, the weights of the atoms are searched, and the sum is made:

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Then, one mole of Fe2O3 has a mass of 159.65 grams.

2- Then, the relationship between the Fe2O3 that will react and the iron to be produced. With the previous calculation, we can say that with one mole of Fe2O3, two moles of Fe can be produced. Passing this relationship to the molar masses, it would be as follows:

1 mole of Fe2O3_____ 2 moles of Fe

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