Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
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)
<span><span>Convert the percentages into decimals (you can do that by dividing the percent by 100), then multiply that by its
corresponding mass to find its relative amount/ contribution to the
atomic mass of chromium. After doing so, add all of the obtained values
together to get the average mass.
</span>
83.79% = .08379
9.50% = .095
4.35% = .0435
2.36% = .0236
Average mass of chromium = 0.8379(51.94) + 0.095( 52.94) + 0.0435(49.95) + 0.0236(53.94)
Answer: 52amu
P.S. never forget units
</span>
They are moving away from each other
Answer:
i mean kinda they should throw people in the air tho lol
Explanation: