First, you have to find now many moles of octane are present in 191.6g of octane. To do this you need to do this you need to divide 191.6g by its molar mass (which is 114g/mol). This will give you 1.681 moles of octane. Then you need to use the fact that 2 moles of octane are us ed to make 16 moles of carbon dioxide to find how many moles of carbon dioxide 1.681mole of octane produces. To do this you need to multiply 1.681mole by 16/2 to get 13.45mol carbon dioxide. The final step is to find the number of grams presswnt in 13.446 moles of carbon dioxide. To do this you need to multiply 13.446 mole by carbon dioxides molar mass (which is 44g/mol) to get 591.6 g of carbon dioxide.
Therefore, 591.6g of carbon dioxide is produced when 191.6 grams of octane is burned.
I hope this helps. Let me know in the comments if anything is unclear.
The answer is 18.02 g/mol
1) Silicon dioxide formula: SiO2 ....... 2 is a subscript for the O atom
2) From the formula you have 1 molecula of SiO2 contains 1 atom of SiO2
3) Then, 0.100 mol of SiO2 contains 0.1 mol of Si
4) Multiply by Avogadro's number: 0.100mol * 6.022*10^23 atoms/mol= 6.02*10^22 atoms
Answer: 6.02*10^22 atoms
Answer: There are 4.375 moles in 2.5 L of 1.75 M 
Explanation:
To calculate the number of moles for given molarity, we use the equation:
Molarity of solution = 1.75 M
Volume of solution = 2.5 L
Putting values in equation , we get:

Answer:
There are 0,89 moles of nitrous oxide gas in the balloon.
Explanation:
We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:
PV= nRT ---> n= PV/RT
n= 1,09 atm x 20,0 L /0.082 l atm / K mol x 298 K
<em>n= 0,89212637 mol</em>