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nlexa [21]
3 years ago
8

Copper reacts with silver nitrate through single replacement.

Chemistry
1 answer:
-Dominant- [34]3 years ago
6 0

Answer:

  • Part a) 0.0104 moles copper(II) nitrate.

  • Part b)

            i) 0.0418 mole Cu

            ii) 0.0209 mol Ag NO₃

Explanation:

<u>1) Balanced chemical reaction (single replacement):</u>

In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)

  • Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag

<u>2) Mole ratio: </u>

  • 1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag

<u />

<u>3) Moles of Ag</u>

  • n = mass in grams / atomic mass
  • atomic mass of Ag: 107.868 g/mol
  • n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag

<u>4) Moles of copper(II) nitrate:</u>

  • Set the proportion using the mole ratio:
  • 2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x
  • Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ =  0.0104 moles Cu(NO₃)₂

That is the answer of part a: 0.0104 moles copper(II) nitrate.

<u>5) Moles of each reactant</u>

i) Cu:

  • Set a proportion using the theoretical mole ratio

        1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag

  • Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu

ii) Ag NO₃

  • Set a proportion using the teoretical mole ratio

   

       2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag

  • Solve for x: x = 0.0209 mol Ag NO₃
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ZanzabumX [31]
Correction: The temperature change is from 20 °C to 30 °C.

Answer:
               Cp  =  1.0032 J.g⁻¹.°C⁻¹

Solution:

The equation used for this problem is as follow,

                                                  Q  =  m Cp ΔT   ----- (1)

Where;
            Q  =  Heat  =  5016 J

            m  =  mass  =  500 g

            Cp  =  Specific Heat Capacity  =  ??

            ΔT  =  Change in Temperature  =  30 °C  -  20 °C  =  10 °C

Solving eq. 1 for Cp,

                                 Cp  =  Q / m ΔT

Putting values,
                                 Cp  =  5016 J / (500 g × 10 °C)

                                 Cp  =  1.0032 J.g⁻¹.°C⁻¹
4 0
3 years ago
Use the table to answer the question
My name is Ann [436]

Answer:

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Explanation:

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When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

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Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

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3 0
2 years ago
[OH-] for a solution is
solong [7]

Answer:

B = basic

Explanation:

Given data:

[OH⁻] = 5.35×10⁻⁴M

pH = ?

Solution:

pOH = -log[OH⁻]

pOH = - [5.35×10⁻⁴]

pOH = 3.272

it is known that,

pH + pOH = 14

pH = 14- pOH

pH = 14 - 3.272

pH = 10.728

The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.

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