Molar mass of butane = 12g/mol*4+1g/mol*10=58g/mol
Mass of 2 moles of butane=2mol*58g/mol=116g
2c4h10(g)+13o2(g)->8co2(g)+10h2o(l)
<span>116g 8 moles</span>
<span>3.20g x</span>
116g butane/8moles CO2=3.20g butane/x
x=3.20g butane*8moles CO2/116g butane=0.2207moles CO2
T=23°C=296K
PV=nRT
V=nRT/P=0.2207moles*0.082(atm*dm³/(K/mol))*296K/(1atm)=5.36dm³
Answer:
The amount of energy transferred to the coin is 28.5 joules.
Explanation:
Have a great rest of your day
#TheWizzer
Then I would guess the answer is D because Na2OH is much more common and stable than Na2O is <span />