<span>cu=1*63g=63g
n=2*14g=28g
o=6*16g=96g
total=187g/mol
122(1 mol/187)(6.02*10^23)= 3.92*10^23 </span><span>
</span>
The question is incomplete. The complete question is :
C. Balance these fossil-fuel combustion reactions. (1 point)
C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat
CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
Solution :
C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat
When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.
CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat
When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.
C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat
When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.
C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat
When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.
Answer: 0.8g/cm^3
Explanation:
In seeing your problem, I see an issue with your units for centimeter. The volume is in the third dimension, so we use cm^3 every single time for the volume. That way, we can arrive to the correct density (in g/cm^3)
Density = mass/volume and so
Density = 8g/10cm^3
= 0.8g/cm^3
Answer:
The object placed in the water has a volume of 19 cm³
Explanation:
<u>Step 1: </u>Data given
volume of the cylinder before adding the object = 28 mL = 28 cm³
After adding an object with volume X the volume rises to 47 mL = 47 cm³
<u>Step 2:</u> Calculate the volume of the object
Volume of the object = Final volume - initial volume
Volume of the object = 47 cm³ (or 47 mL) - 28 cm³ ( or 28 mL) = 19 cm³ (or 19 mL)
The object placed in the water has a volume of 19 cm³
