I'm pretty sure it's B) Mass
Answer:
The amount of moles of Fe in 5.22*10²² atoms of Fe is 0.0867.
Explanation:
Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension, so it is considered a pure number that allows describing a physical characteristic without an explicit dimension or unit of expression. Avogadro's number applies to any substance.
Then you can apply the following rule of three: if 6.023*10²³ atoms are contained in 1 mole, 5.22*10²² atoms are contained in how many moles?

amount of moles= 0.0867
<u><em>The amount of moles of Fe in 5.22*10²² atoms of Fe is 0.0867.</em></u>
Answer:
.079 moles of Nirogen gas (N2)
Explanation:
You can see from the equaton that each ONE mole of N2 produces TWO moles of NH3.
Find the number of moles of NH3 produced.
Using Periodic Table : Mole wt of NH3 = 17 gm/mole
2.7 gm / 17 gm/mole = .1588 moles
One half as many moles of N2 are needed = .079 moles
Answer:
The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is<u> 394.91 K.</u>
Explanation:
The formula for molal boiling Point elevation is :

= elevation in boiling Point
= Boiling point constant( ebullioscopic constant)
m = molality of the solution
<em>i =</em> Van't Hoff Factor
Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .
In solution Mg3(PO4)2 dissociates as follow :

Total ions after dissociation in solution :
= 3 ions of Mg + 2 ions of phosphate
Total ions = 5
<em>i =</em> Van't Hoff Factor = 5
m = 8.5 m
= 0.512 °C/m
Insert the values and calculate temperature change:



Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

= 373.15 K[/tex]
21.76 = T - 373.15
T = 373.15 + 21.76
T =394.91 K
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.