If you look at the periodic table, underneath each element it give you the atomic mass, which has the units g/mol. So CO2 have a molecular mass (MM) of those combined atomic masses.
The MM for CO2 would be...
12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
This can be used as a conversion to change from moles to grams.
8 mol * (44.01 g / 1 mol) = 352.08 g or 6 x 10 to the 2 g for the significant figures (sig figs) at are shown in your problem.
You may have just typed 8 instead of 8.0 or 8.00 which would change your sig figs. It is all important to right now all your zeros, since almost all chemistry calculations involve significant figures.
I solved the above equation using dimensional analysis and had to cancel out the units, in case this is your first chemistry class.
I hope I helped!
Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
= 
u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u
Answer:
The reaction rate k is 0.0012563 (1/hour).
Explanation:
We considered the reactions occurring in the plant as first order, and represented by this equation:
where y is the BOD at time t, L is the initial value of BOD and k is the reaction rate.
If we replaced with the values
y = 2 mg O2/l (1% of the initial value)
L = 200 mg 02/l
t = 8 hr
We can calculate k
The reaction rate k is 0.0012563 1/hour.
Answer:
No
Explanation:
One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.
When one gram phosphorus and 6 gram of iodine react they gives 8.234 g
ram of PI₃ .
Given data:
Mass of phosphorus = 1 g
Mass of iodine = 6 g
Mass of PI₃ = ?
Solution:
Chemical equation:
P₄ + 6I₂ → 4PI₃
Number of moles of P₄:
Number of moles = Mass /molar mass
Number of mole = 1 g / 123.9 g/mol
Number of moles = 0.01 mol
Number of moles of I₂:
Number of moles = Mass /molar mass
Number of moles = 6 g / 253.8 g/mol
Number of moles = 0.024 mol
Now we will compare the moles of PI₃ with I₂ and P₄.
I₂ : PI₃
6 : 4
0.024 :
4/6×0.024 = 0.02
P₄ : PI₃
1 : 4
0.01 : 4 × 0.01 = 0.04 mol
The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.
Mass of PI₃ = moles × molar mass
Mass of PI₃ = 0.02 mol × 411.7 g/mol
Mass of PI₃ = 8.234 g
