<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>
Answer:
measuring the consumption of the products and the rate of conversion of products to reagents
Explanation:
It must be taken into account that the reaction rate also depends on the presence of a catalyst, since these advance the rate of the reaction.
A = Z + n
A = 12 + 10
A = 22
Answer C
hope this helps!
Answer:
Have some attraction towards each other
Explanation:
Gases deviate from the ideal gas behavior because their molecules have forces of attraction between them. At high pressure, the molecules of gases are very close to each other so the molecular interactions start operating and these molecules do not strike the walls of the container with full impact.
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Answer:
Explanation:
From the given information:
The density of O₂ gas = 
here:
P = pressure of the O₂ gas = 310 bar
= 
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴

= 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:

where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:



After solving;
V = 0.1152 L
∴

= 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.