Answer:
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Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
<span>STP means standard temperature
and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown
gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the
numer of moles and can be represented as mass of the gas, m, divided by the
molar mass, c. so we have,</span>
PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P
substitute the values into the equation,
c = [(0.63g/L)(0.08206
L-atm/mol-K)(273K)]/(1atm)
<u>c = 14.11 g/mol</u>
I dont know if this correct, but i hope it help
Pretty sure it’s 2, increase in strength of hurricanes
