Answer:
![\boxed {\boxed {\sf B. \ 22 \ grams}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20B.%20%5C%2022%20%5C%20grams%7D%7D)
Explanation:
We need to use the formula for heat of vaporization.
![Q=H_{vap}*m](https://tex.z-dn.net/?f=Q%3DH_%7Bvap%7D%2Am)
Identify the variables.
- The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
- Q is the energy, in this problem, 50,000 Joules.
- m is the mass, which is unknown.
![H_{vap}=2260 \ J/g\\Q=50,000 \ J \\](https://tex.z-dn.net/?f=H_%7Bvap%7D%3D2260%20%5C%20J%2Fg%5C%5CQ%3D50%2C000%20%5C%20J%20%5C%5C)
Substitute the values into the formula.
![50,000 \ J=2260 \ J/g*m](https://tex.z-dn.net/?f=50%2C000%20%5C%20J%3D2260%20%5C%20J%2Fg%2Am)
We want to find the mass. We must isolate the variable, m.
m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.
![\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}](https://tex.z-dn.net/?f=%5Cfrac%7B50%2C000%20%5C%20J%7D%7B2260%20%5C%20J%2Fg%7D%20%3D%5Cfrac%7B2260%20%5C%20J%2Fg%2Am%7D%7B2260%20%5C%20J%2Fg%7D)
![\frac{50,000 \ J}{2260 \ J/g} =m](https://tex.z-dn.net/?f=%5Cfrac%7B50%2C000%20%5C%20J%7D%7B2260%20%5C%20J%2Fg%7D%20%3Dm)
Divide. Note that the Joules (J) will cancel each other out.
![\frac{50,000 \ }{2260 \ g} =m](https://tex.z-dn.net/?f=%5Cfrac%7B50%2C000%20%5C%20%7D%7B2260%20%5C%20g%7D%20%3Dm)
![22.1238938 \ g =m](https://tex.z-dn.net/?f=22.1238938%20%5C%20g%20%3Dm)
Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.
![22 \ g \approx m](https://tex.z-dn.net/?f=22%20%5C%20g%20%5Capprox%20m)
The mass is about 22 grams, so choice B is correct.
Natural, Organic & Manmade Fibers. after flame is out there is a soft gray ash!
The formula to determine the average atomic mass is:
Average atomic mass =
+
+ ........
-(1)
The atomic mass of first isotope of magnesium = 24 u (given)
The natural abundance of first isotope = 78.70 % (given)
The atomic mass of second isotope of magnesium = 25 u (given)
The natural abundance of first isotope = 10.13 % (given)
The atomic mass of third isotope of magnesium = 26 u (given)
The natural abundance of first isotope = 11.17 % (given)
Substituting the values in formula (1):
atomic weight = ![(24 \times 0.7870) + (25 \times 0.1013) + (26 \times 0.1117)](https://tex.z-dn.net/?f=%2824%20%5Ctimes%200.7870%29%20%2B%20%2825%20%5Ctimes%200.1013%29%20%2B%20%2826%20%5Ctimes%200.1117%29)
atomic weight = ![24.4625 u](https://tex.z-dn.net/?f=24.4625%20u)
Hence, the atomic weight of naturally occurring isotopic mixture is
.