Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
Answer:
A = 1,13x10¹⁰
Ea = 16,7 kJ/mol
Explanation:
Using Arrhenius law:
ln k = -Ea/R × 1/T + ln(A)
You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).
Using the values you will obtain:
y = -2006,9 x +23,147
As R = 8,314472x10⁻³ kJ/molK:
-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹
<em>Ea = 16,7 kJ/mol</em>
Pre-exponential factor is:
ln A = 23,147
A = e^23,147
<em>A = 1,13x10¹⁰</em>
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I hope it helps!
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