Answers:
A) 2040 kg/m³; B) 58 600 km
Explanation:
A) Density


<em>B) Radius</em>



![r= \sqrt [3]{ \frac{3V }{4 \pi } }](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5B3%5D%7B%20%5Cfrac%7B3V%20%7D%7B4%20%5Cpi%20%7D%20%7D)
![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}](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5B3%5D%7B%20%5Cfrac%7B3%5Ctimes%208.268%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%7B%20m%7D%5E%7B3%7D%7D%7B4%20%5Cpi%20%7D%20%7D%3D%20%5Csqrt%20%5B3%5D%7B%201.974%20%5Ctimes%2010%5E%7B23%7D%20%5Ctext%7B%20m%7D%5E%7B3%7D%7D%3D%205.82%20%5Ctimes%2010%5E%7B7%7D%20%5Ctext%7B%20m%7D%3D%5Ctext%7B58%20200%20km%7D)
Answer:
The percent yield of this reaction is 84.8 % (option A is correct)
Explanation:
Step 1: Data given
The student isolated 15.6 grams of the product = the actual yield
She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams
Step 2: Calculate the percent yield
Percent yield = (actual yield / theoretical yield ) * 100 %
Percent yield = (15.6 grams / 18.4 grams ) * 100 %
Percent yield = 84.8 %
The percent yield of this reaction is 84.8 % (option A is correct)
Mg + 1/2 O2 → MgO
1 mol = 24 g of Mg
X mol = 12 g of Mg
x = 0.5 moles of Mg
Mg :MgO = 1:1 (coefficient from equations using mole ratio)
So
0.5 moles of MgO
1 mol MgO = (24+16) g = 40 g
0.5 moles of MgO = 0.5 × 40
= 20 g of MgO produced