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8090 [49]
3 years ago
12

Can someone please help me with these 2 questions?

Chemistry
1 answer:
nydimaria [60]3 years ago
4 0
<h3>1. Answer:</h3>

2MnO4−(aq) + 5C2O42−(aq) + 16 H3O+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 24 H2O(l)

  • Element being reduced -Mn
  • Element being oxidized - C
<h3>Explanation:</h3>
  • Redox reactions are reactions that involves both oxidation and reduction.
  • Reduction involves the gain of electrons by an atom while oxidation is the loss of electrons.
  • A reduction may also refer to a decrease in the oxidation number of an element while oxidation is an increase in the oxidation number of an element in a reaction.
  • In the Equation:

2MnO4−(aq) + 5C2O42−(aq) + 16 H3O+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 24 H2O(l)

  • The oxidation number of Mn on the left side of the equation is +7 and on the right side of the equation is +2, therefore Mn has undergone reduction.
  • On the other hand, the oxidation number of Carbon (C) on the left side of the equation is +3 and on the right side of the equation is +4, therefore Carbon(C) has undergone oxidation.
<h3>2. Answer</h3>

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(aq) + 4 H2O(l)

  • Element being reduced - N
  • Element being oxidized - Cu

<h3>Explanation </h3>
  • In the Equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(aq) + 4 H2O(l)

  • The oxidation number of Cu on the left side is 0 and on the right side is +2, therefore, copper has undergone oxidation.
  • The oxidation number of Nitrogen on the left side is +5 and on the right side is +2, therefore, nitrogen has undergone reduction.

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Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of 0.100
yulyashka [42]

Answer:

ΔH = -59.6kJ/mol

Explanation:

The reaction that occurs between Ag⁺ and Cl⁻ ions is:

Ag⁺ + Cl⁻ → AgCl(s) + ΔH

To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.

<em>Moles of reaction:</em>

Moles of Ag⁺ and Cl⁻ added are:

Ag⁺: 0.100L * (0.100mol / L) = 0.01moles

Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles

That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles

<em>Heat released:</em>

To find heat released we must use coffe cup calorimeter equation:

Q = C*m*ΔT

<em>Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).</em>

Replacing:

Q = C*m*ΔT

Q = 4.18J/g°C * 200g * 0.70°C

Q = 585,2J

Is total heat released.

The calorimeter absorbs:

15.5J / °C * 0.7°C = 10.85

Thus, when 0.01 moles reacts, 585.2J + 10.85  = 596.05J are released (Heat released is heat abosrbed by calorimeter + Heat absorbed by water) and ΔH is:

ΔH = 596.05J / 0.01 moles =

ΔH = 59605J / mol =

<h3>ΔH = -59.6kJ/mol</h3>

<em>As heat is released, ΔH < 0.</em>

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