Answer:
Equilibrium constant for 
is 0.5
Equilibrium constant for decomposition of 
is 
Explanation:
dissociates as follows:
initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for is 0.5
Now calculate the equilibrium constant for decomposition of
It is given that 
is decomposed.
decomposes as follows:
initial 1.0 M 0 0
at eq. concentration of is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
![[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D3.3%5Ctimes%2010%5E%7B-5%7D%20%3D%203.3%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of is
The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH
Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction
Answer: The questions looks unclear
Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.
1. D belongs to group 4
E. Belongs to group 7
B belongs to group 1
A belongs to group 8. A noble gas.
R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8
Answer:
Mole fraction of nitrogen = 0.52
Explanation:
Given data:
Temperature = 31.2 °C
Pressure = 870.2 mmHg
Volume = 15.1 L
Mass of mixture = 24.1 g
Mole fraction of nitrogen = ?
Solution:
Pressure conversion:
870.2 /760 = 1.12 atm
Temperature conversion:
31.2 + 273 = 304.2 K
Total number of moles:
PV = nRT
n = PV/RT
n = 1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K
n = 16.9 L.atm. /25 L.atm. mol⁻¹
n = 0.676 mol
Number of moles of nitrogen are = x
Then the number of moles of CO₂ = 0.676 - x
Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)
24.1 = 28x + ( 29.7 -44x)
24.1 - 29.7 = 28x - 44x
-5.6 = -16 x
x = 0.35
Mole fraction of nitrogen:
Mole fraction of nitrogen = moles of nitrogen / total number of moles
Mole fraction of nitrogen = 0.35 mol / 0.676 mol
Mole fraction of nitrogen = 0.52
Answer:
The solubility of X in water at 17°C is 0.110 g/mL.
Explanation:
The water of a rock pool lined with mineral crystals is a <em>saturated solution</em> of said mineral, this means the concentration of X in those 36 mL is the solubility of compound X in water at 17 °C.
- This means<u> it is possible to calculate said solubility</u>.
The dilution of the sample is not relevant, nor is that 500 mL volume. What's important is that 3.96 g of X form a saturated solution with 36.0 mL of water, so the solubility is:
- 3.96 g / 36.0 mL = 0.110 g/mL
