The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
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Answer:
C. involve the attraction of opposite charges
Explanation:
<em>Ionic bonding</em> involves the attraction between <em>oppositely charged ions</em>, as in Na⁺ Cl⁻.
<em>Covalent bonding</em> involves the attraction between <em>negatively charged electrons and positivey charged nuclei</em>, as in a C-H bond.
A is <em>wrong</em>. Ionic bonding involves the transfer of electrons.
B is <em>wrong</em>. Covalent bonding involves the sharing of electrons.
D is <em>wrong</em>. Ionic bonds are usually stronger than covalent bonds.
The concentration of diluted solution is 0.16 M
<u>Explanation:</u>
As, the number of moles of diluted solution and concentrated solution will be same.
So, the equation used to calculate concentration will be:
where,
are the molarity and volume of the concentrated HCl solution
are the molarity and volume of diluted HCl solution
We are given:
Putting values in above equation, we get:
Hence, the concentration of diluted solution is 0.16 M
Answer: The approximate equilibrium partial pressure of 
is 3.92 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,
![K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
![1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
On reversing the reaction:
initial pressure 4.00atm 2.00 atm 0
eqm (4.00-2x)atm (2.00-x) atm 2x atm
![K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)
![0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}](https://tex.z-dn.net/?f=0.67%5Ctimes%2010%5E5%3D%5Cfrac%7B2x%5D%5E2%7D%7B%5B4.00-2x%5D%5E2%5Ctimes%20%5B2.00-x%5D%7D)
![[H_2S]=2x=2\times 1.96=3.92 atm](https://tex.z-dn.net/?f=%5BH_2S%5D%3D2x%3D2%5Ctimes%201.96%3D3.92%20atm)
Thus approximate equilibrium partial pressure of is 3.92 atm
