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padilas [110]
3 years ago
5

What is the [pb2+] in a solution made by adding 100 g of pbcl2(s) (mm = 278.1 g/mol) to a 5.4 m solution of nacl and allowing it

to come to equilibrium? ignore any changes to solution volume due to the addition of pbcl2(s). once solution reaches equilibrium, it is noticed that some solid pbcl2 remains undissolved. the ksp for pbcl2 is 1.6 × 10−5?
Chemistry
1 answer:
Evgen [1.6K]3 years ago
7 0
                    PbCl₂ ⇄ Pb²⁺ + 2 Cl⁻
Initially                         0         5.4 M
Change          -x         + x        + 2x
Equilibrium                  x          5.4 + 2x

Ksp = [Pb²⁺] [Cl⁻]²
1.6 x 10⁻⁵ = x (5.4 + 2x)²
since x <<<< 5.4 so 2x + 5.4 = 5.4
1.6 x 10⁻⁵ = x (5.4)²
x = 5.48 x 10⁻⁷
[Pb²⁺] = 5.48 x 10⁻⁷ M
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miskamm [114]

Answer:

High concentration of glutathione should be included in the elution buffer for the given experiment to remove protein from the column.

Explanation:

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7 0
3 years ago
And electro chemical cell has the following standard cell notation:
pochemuha

2Ag⁺(aq) + Mg(s)→ 2Ag(s) + Mg²⁺ (aq)

<h3>Further explanation</h3>

Given

Standard cell notation:

Mg(s) | Mg2+ (aq) || Ag+(aq)| Ag(s)

Required

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Solution

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In reaction:  

Ag⁺ + Mg → Ag + Mg²⁺  

half-reactions

  • at the cathode (reduction reaction)

Ag⁺ (aq) + e⁻ ---> Ag (s)  x2

2Ag⁺ (aq) + 2e⁻ ---> 2Ag (s)

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a balanced cell reaction

<em>2Ag⁺(aq) + Mg(s)→ 2Ag(s) + Mg²⁺ (aq) </em>

5 0
3 years ago
Read 2 more answers
90 POINTSSSS!!!
Katarina [22]

Answer:

\large \boxed{\text{-2043.96 kJ/mol}}

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn:    C₃H₈(g) +   O₂(g) ⟶   CO₂(g) +   H₂O(g)

Balanced eqn:    C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

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\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}

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\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\=  \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}

ΔᵣH° is negative, so the reaction is exothermic.

5 0
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Answer:

same I hate all the frickin ads ;-;

you gave a good day too mate

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