Hi Sydney!
I can't draw in this question, but there is a picture showing this phase for you to follow when you draw it.
Hope This Helps :)
Answer:
The empirical formula is the simplest form;
Given:
Oxygen O at 94.1% and
H at 5.9%
Assume 100grams.
94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O
5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H
There is one mole of O for each mole of H so the empirical formula is 
and written as OH.
Answer: 122 moles
Procedure:
1) Convert all the units to the same unit
2) mass of a penny = 2.50 g
3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).
=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.
4) find how many times the mass of a penny is contained in the mass of the Moon.
You have to divide the mass of the Moon by the mass of a penny
7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies
That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.
5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23
7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.
Answer: 122 mol
Answer:
25.42 atm
Explanation:
Data Given:
Volume of a gas ( V )= 2.00 L
temperature of a gas ( T ) = 310 K
number of moles (n) = 2 mol
Pressure of a gas ( P ) = to be find
Solution:
Formula to be used
PV= nRT
Rearrange the above formula
P = nRT / V . . . . . . . . . . (1)
Where R is ideal gas constant
R = 0.08205 L atm mol⁻¹ K⁻¹
Put values in equation 1
P = nRT / V
P = 2 mol x 0.08205 L atm mol⁻¹ K⁻¹ x 310 k / 2 L
P = 50.84 L atm / 2 L
P = 25.42 atm
P ressure of gas (P) will be = 25.42 atm
All problems are caused by a factor in which human beings play an important role in its cause.
