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mars1129
2 days ago
12

Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are

mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 ∘C to 27.7 ∘C.
a) Before mixing, how many grams of Cu are present in the solution of CuSO4?
b) Predict the identity of the precipitate in the reaction.
c) Write complete equation for the reaction that occurs when the two solutions are mixed.
d) Write net ionic equation for the reaction that occurs when the two solutions are mixed.
e) From the calorimetric data, calculate ΔH for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, and that the specific heat and density of the solution after mixing are the same as that of pure water.
Chemistry
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An atom of the element zinc has an atomic number of 30 and a mass number of 65. How many protons does an uncharged zinc atom hav
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Answer: The correct option is, 30 protons

Explanation:

Element = Zinc

Atomic number = 30

Atomic mass number = 65

Recall that the atomic number equals the number of electrons and protons.

Atomic number = Number of electrons = Number of protons = 30

Number of neutrons = Atomic mass - Number of protons = 65 - 30 = 35

Therefore, an uncharged zinc atom has 30 protons.

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Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c
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Answer:

[Cl⁻] = 0.016M

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To begin, we analyze the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This indicates a solubility equilibrium, resulting in the formation of lead(II) chloride precipitate. The salt can dissociate as follows:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

Given that this is an equilibrium scenario, the Kps serves as the constant (Solubility product):

Kps = s. (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵. 1/4)

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