Answer:
0.26×10²³ molecules
Explanation:
Given data:
Volume of gas = 1.264 L
Temperature = 168°C
Pressure = 946.6 torr
Number of molecules of gas = ?
Solution:
Temperature = 168°C (168+273= 441 K)
Pressure = 946.6 torr (946.6/760 = 1.25 atm)
Now we will determine the number of moles.
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.25 atm ×1.264 L / 0.0821 atm.L/ mol.K ×441 K
n = 1.58 /36.21 /mol
n = 0.044 mol
Now we will calculate the number of molecules by using Avogadro number.
1 mol = 6.022×10²³ molecules
0.044 mol × 6.022×10²³ molecules/ 1mol
0.26×10²³ molecules
The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
The equation for aqueous solution of H₂CO₃ is
H₂CO₃ → H₂O + CO₂
The charge balance equation is
[HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]
Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
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<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:



Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:

<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:

Hence, the value of equilibrium constant for the given reaction is 56.61
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