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)
The answer is B) gain 8 electrons
Answer:
Pentan-2-ol
Explanation:
On this reaction, we have a <u>Grignard reagent</u> (ethylmagnesium bromide), therefore we will have the production of a <u>carbanion</u> (step 1). Then this carbanion can <u>attack the least substituted carbon</u> in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the <u>treatment with aqueous acid</u>, when we add acid the <u>hydronium ion</u> (
) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be <u>attacked by the negative charge</u> produced in the second step to produce the final molecule: <u>"Pentan-2-ol".</u>
See figure 1
I hope it helps!
<span>The outer electrons are not as tightly bound as ones closer to the nucleus</span>
The sample of smoke described above can be described as a heterogeneous mixture. This type of mixture do not have uniform properties and composition. So, getting a certain small sample would not represent the whole mixture since it does not have uniform composition.