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madreJ [45]
4 years ago
13

What would be the resulting molarity of a solution made by dissolving 21.9 grams of KCl in enough water to make an 869-millilite

r solution? Show all of the work needed to solve this problem.
Chemistry
1 answer:
viktelen [127]4 years ago
8 0
Molar mass KCl = <span>74.5513 g/mol

Number of moles:

21.9 / 74.5513 => 0.293 moles

Volume = 869 mL / 1000 => 0.869 L

Molarity = moles / Volume

Molarity = 0.293 / 0.869

=> 0.337 M</span>
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When the reaction 3 no(g) → n2o(g) + no2(g) is proceeding under conditions such that 0.015 mol/l of n2o is being formed each sec
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Answer:

1) The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.

2) The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt  = 0.045 mol/L.s.

Explanation:

  • For the reaction:

<em>3NO(g) → N₂O(g) + NO₂(g).</em>

The rate of the reaction = -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt = Δ[NO₂]/Δt.

Given that: Δ[N₂O]/Δt = 0.015 mol/L.s.

<em>1) The rate of the overall reaction is?</em>

The rate of the overall reaction = Δ[N₂O]/Δt = 0.015 mol/L.s.

<em>2) The rate of change for NO is?</em>

The rate of change for NO = - Δ[NO]/Δt.

∵ -1/3 Δ[NO]/Δt = Δ[N₂O]/Δt.

<em>∴ The rate of change for NO = - Δ[NO]/Δt = 3 Δ[N₂O]/Δt </em>= 3(0.015 mol/L.s) = <em>0.045 mol/L.s.</em>

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When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)
alexandr402 [8]

The question is incomplete, the complete question is;

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.

Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

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3.81 g of lead

Explanation:

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