Answer:
The molecular formula of estradiol is: 
.
Explanation:
Molar mass of of estradiol = M= 272.37 g/mol
Let the molecular formula of estradiol be 
Percentage of an element in a compound:
Percentage of carbon in estradiol :
x = 18.0
Percentage of hydrogen in estradiol :
y = 24.2 ≈ 24
Percentage of oxygen in estradiol :
z = 2
The molecular formula of estradiol is: 
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is
% by mass = (mass of the solute / mass of the solution)*100
= mass of solute / (mass of the solute + mass of the solvent)*100
= (27.7 g CaCl2 / 27.7g + 375g) * 100
= 6.88
The answer is "elements" :)
First. moles is just a label for a number of things. just like a
dozen = 12, a gross = 144, a mole = 6022 with another 20 zeros after the
2
next
moles = mass / molecular weight.
molecular weight = sum of atomic mass from the periodic table
atomic mass MnO2 = atomic mass Mn + 2 x atomic mass O
= 54.94 + 2 x 16 = 86.94 g/mole
so moles MnO2 = 98.0 grams / (86.94 g/mole) = 1.13 moles
notice that I only gave 3 digits? that because of sig figs read the link below if you don't understand....
mw C5H12 = 5 x 12 + 12 x 1 = 72 g/mole
so moles C5H12 = 12.0 g / 72.0 g/mole = 0.167 moles
mw XeF6 = 131.3+ 6 x 19.00 = 245.3
so moles XeF6 = 100 g / 245.3 g/mole = 0.4077 moles
I've also provided a link to a periodic table. if you need atomic weights click on any element and it will give you the
details.
Low electron affinity and large atomic radius
Metallic character decreases across a period (from left to right) and increases down a group.
Electron affinity increases from left to right within a period. This is caused by the decrease in atomic radius. Electron affinity decreases from top to bottom within a group. This is caused by the increase in atomic radius.
