Hey there,
<span>A scientist triples the temperature of a gas at constant pressure. By what factor does the gas’s volume change?
I believe that when the scientist triples the temperature of the gas at a certain and constant pressure, the factor of the gas volume would change most likely with the double amount of pressure that the scientist put in the first place.
~Jurgen</span>
According to the balanced equation of this reaction:
N2(g) + 3H2(g) ↔ 2NH3(g)
and when we have Kp = 4.51 x 10^-5 so, in the Kp equation we will substitute by the value of the P for each gas to compare the value with Kp = 4.51x10^-5
a) when we have 98 atm NH3, 45 atm N2, 55 atm H2 by substitution in Kp equation:
Kp= [p(NH3)]^2 / [p(N2)]*[p(H2)]^3 = [98]^2 / [45]*[55]^3
= 1.28x10^-3
So here the value is higher than the value of the given Kp.
so the reaction will go leftwards toward the reactants ( to reduce the value of Kp) to reach the equilibrium.
b) When 57 atm NH3, 143 atm N2, No H2 so like a) by substitution:
Kp = [57]^2 / [143] = 22.7
So the reaction will go leftwards toward the reactants to reduce the value of Kp to reach equilibrium.
c) when 13 atm NH3, 27 atm N2, 82 H2
Kp = [13]^2 / [27]*[82]^3 = 1.135 x 10^-5 So this value is lower than the Kp which is given.
so, the reaction will go towards the right toward the products to increase the value of Kp to reach the equilibrium.
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Answer:
Explanation:
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
PV =nRT
at STP P= 1atm. T= 273 K
n is the number of moles. O2 has a molar mass of 32.
30 gm of O2 is 30/32= 0.94 =n
PV = nRT
at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K
SO
1 X V = 0.94 X 0.082 X 273
using high school freshman algebra,
V= 0.94 X 0.082 X 273 = 21L
using high school algebra I,
V=
Answer: The number of grams of
in 1620 mL is 1.44 g
Explanation:
According to ideal gas equation:

P = pressure of gas = 1 atm (at STP)
V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)
n = number of moles = ?
R = gas constant =
T =temperature =


Mass of hydrogen =
The number of grams of
in 1620 mL is 1.44 g