First you should know that there is seven oxygen atoms in one Mn2O7
So
2.00 moles of Mn2O7 contain 14.00 moles of oxygen...
Then you multiply this no. with Avagadro no....
from formula
Number of moles= no. of particles/avagadro's no..
14.00×6.02×10²³=84.28 atoms of oxygen...
Answer:
516.77 grams of Argon gas is present
Explanation:
Using the gas formula
PV = nRT
number of moles (n) = mass / molar weight or mass
P = pressure = 3.4 atm
V = volume = 72 L
R = gas constant = 0.082 L atm mol^-1 K^-1
T = temperature = 225 K
MM = molar mass of Ar = 38.984 g/mol
PV = mRT/ MM
m = PV MM / RT
m = 3.4 * 72 * 38.948 / 0.082 * 225
m = 9534.4704 / 18.45
m = 516.77 grams
the mass of Ar gas you have is 516.77 grams.
Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)
Hydrogen is usually –1. This is INCORRECT. The oxidation number for H is +1.
Oxygen is usually –2. This is CORRECT.
A pure group 1 element is +1. This is INCORRECT. It does not follow. This will depend on the other elements and the overall charge.
A monatomic ion is 0. This is INCORRECT. Diatomic ion is 0.
