Answer: 72.93 litres
Explanation:
Given that:
Volume of gas (V) = ?
Temperature (T) = 24.0°C
Convert 24.0°C to Kelvin by adding 273
(24.0°C + 273 = 297K)
Pressure (P) = 1.003 atm
Number of moles (n) = 3 moles
Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1
Then, apply ideal gas equation
pV = nRT
1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K
1.003 atm•V = 73.15 atm•L
Divide both sides by 1.003 atm
1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm
V = 72.93 L
Thus, the volume of the gas is 72.93 litres
- C_5H_8+13/2O_2—»5CO_2+4H_2O
Balanced one
- 2C_5H_8+13O_2—»10CO_2+8H_2O
Moles of Pentyne
- Given mass/Molarmass
- 34/68
- 0.5mol
Moles of H_2O
1mol releases 241.8KJ
2mol releases 241.8(2)=483.6KJ
<span> iron (Fe), ruthenium (Ru), osmium (Os) and hassium (Hs). They are all transition metals.</span>
Answer: 50.7 grams
Explanation:
To calculate the moles, we use the equation:
a) moles of 
![\text{Number of moles}=molarity\times {\text {Volume in L]}=0.417M\times 0.528L=0.220moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3Dmolarity%5Ctimes%20%7B%5Ctext%20%7BVolume%20in%20L%5D%7D%3D0.417M%5Ctimes%200.528L%3D0.220moles)
The balanced chemical equation is:
is the limiting reagent as it limits the formation of product and 
is in excess.
According to stoichiometry :
2 moles of give = 1 mole of 
Thus 0.220 moles of give=
of
Mass of 
Thus 50.7 g of will be formed.
