Answer:
0.52 L.
Explanation:
Let P be the initial pressure.
From the question given above, the following data were obtained:
Initial pressure (P1) = P
Initial volume (V1) = 1.04 L
Final pressure (P2) = double the initial pressure = 2P
Final volume (V2) =?
The new volume (V2) of the gas can be obtained by using the the Boyle's law equation as shown below:
P1V1 = P2V2
P × 1.04 = 2P × V2
1.04P = 2P × V2
Divide both side by 2P
V2 = 1.04P /2P
V2 = 0.52 L
Thus, the new volume of the gas is 0.52 L.
4Al(s) + 3O2(g) --> 2Al2O3(s) This is the balanced.
From the equation:
4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3
3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3 (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3
4.5 moles of O2 reacted with 6moles of Al to produce 3moles of Al2O3
(3) is the answer. 6 mol of Al.
the reaction is
2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)
Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2
Given
moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062
moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012
moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019
moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138
Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54
