Explanation:
2CO (g) + O2 (g) → 2CO2 (g) ∆H = –566.0 kJ exothermic
2SO2(g) + O2(g) <=> 2SO3(g). Delta H = -198.2 kJ/mol. exothermic
Answer:
1 mol
Explanation:
Using the general gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in the question;
V = 22.4L
T = 273K
P = 1 atm
R = 0.0821 Latm/molK
n = ?
Using PV = nRT
n = PV/RT
n = (1 × 22.4) ÷ (0.0821 × 273)
n = 22.4 ÷ 22.4
n = 1mol
Answer: The molar volume of any gas at standard pressure and standard temperature is 22.4 liters per mole.
Explanation:The ideal gas law is PV=nRT
P is pressure and if we consider standard pressure, then we have 1.00 atm.
V is volume and that is what we are trying to solve.
n is the number of moles, which is 1.00 moles since we are trying to determine the volume of a gas in one mole.
R is the ideal gas constant which equals
0.0821 (Liters x atmospheres)/(mole x kelvin)
T is the standard temperature which is 273 kelvin.
Rearrange the equation to solve for volume.
V = nRT/P
V = (1.00 mol)(0.0821 L atm/mol K)(273 K)/ 1.00 atm
V = 22.4 L
The given compounds are alkanes. This means that the formula
is C_nH_2n+2
The general trend for boiling point in alkanes is that the
greater number of carbon atoms, the higher the boiling point. If the alkane is
branched, the branched alkane will have a higher boiling point than the
straight chain.
Taking these into account: propane < heptane <
triptane < hexadecane < paraffin
95.611 g/mol that's the answers
