Answer:
1 mol
Explanation:
Using the general gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in the question;
V = 22.4L
T = 273K
P = 1 atm
R = 0.0821 Latm/molK
n = ?
Using PV = nRT
n = PV/RT
n = (1 × 22.4) ÷ (0.0821 × 273)
n = 22.4 ÷ 22.4
n = 1mol
Explanation:
The given reaction is as follows.

Initial : 0.160 0.160 0
Change : -x -x 2x
Equilibrium: 0.160 - x 0.160 - x x
It is given that
= [0.160 - x] = 0.036 M
and,
= [0.160 - x] = 0.036 M
so, x = (0.160 - 0.036) M
= 0.124 M
As, [HI] = 2x.
So, [HI] = 
= 0.248 M
As it is known that expression for equilibrium constant is as follows.
![K_{eq} = \frac{[HI]^{2}}{[H_{2}][I_{2}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5BI_%7B2%7D%5D%7D)
= 
= 47.46
Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.
Answer:
the pressure (torr) of theH₂ gas is 736.2 torr
Explanation:
Given that
The H₂ gas produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.
So; the total pressure in the eudiometer = Pressure in the H₂ gas - Pressure of the water vapor
Where;
= total pressure in the eudiometer = 760.0 torr
= Pressure in the H₂ gas = ???
= Pressure in the water vapor = 23.8 torr
Now:
=
+
-
= +
-
= -
+
= (- 23.8 + 760) torr
= 736.2 torr
Thus; the pressure (torr) of theH₂ gas is 736.2 torr
The bonds of a glucose molecule store chemical energy
Answer:
Percent yield = 65.2 %
Explanation:
The reaction determines this:
2 moles of Al react to 3 moles of chlorine gas in order to produce 2 mol of aluminum chloride.
We determine the moles of chlorine:
52 g . 1mol /70.9g = 0.733 moles of Cl₂
3 moles of gas can produce 2 moles of salt
Then, our 0.733 moles may produce (0.733 . 2) / 3 = 0.489 moles of AlCl₃
We convert moles to mass to predict the 100 % yield
0.489 mol . 133.3 g /mol = 65.19 g
Percent yield = (Yield produced / 100 % yield) . 100
Percent yield = (42.5 g / 65.19g) . 100 = 65.2 %