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GenaCL600 [577]
4 years ago
5

Name the type of equation shown here: 4C5H9O + 27O2 ---> 20CO2 + 18H2O

Chemistry
1 answer:
BartSMP [9]4 years ago
8 0
The correct answer is a Combustion Reaction.
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The first one is false the second one is the green house effect
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3 years ago
17. D i don’t know if i’m writing it wrong or don’t know how to balance but ye
leva [86]

Answer:

2 KL + Pb(NO3)2 = 2 KNO3 + PbL2

Explanation:

KL + Pb(NO3)2 = KNO3 + PbL2 is unballanced equation, just balance from there :)

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3 years ago
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3 years ago
What gets reduced in an electrolytic cell made with nickel and copper electrodes?
jekas [21]

Answer:

D. Ni²⁺  

Explanation:

We know at once that the answer cannot be A or C, because Ni and Cu are already in their lowest oxidation states.

The correct answer must be either B or D.

An electrolytic cell is the opposite of a galvanic cell. In the former, the reaction proceeds spontaneously. In the latter, you must force the reaction to occur.  

One strategy to solve this problem is:

  1. Look up the standard reduction potentials for the half reaction·
  2. Figure out the spontaneous direction.
  3. Write the equation in the reverse direction.

1. Standard reduction potentials

                                E°/V

Cu²⁺ + 2e⁻ ⟶ Cu; 0.3419

Ni²⁺ + 2e⁻ ⟶ Ni;  -0.257

2. Galvanic Cell

We reverse the direction of the more negative half cell and add.

                                       <u>E°/V </u>

Ni ⟶ Ni²⁺ + 2e⁻;           0.257

<u>Cu²⁺ + 2e⁻ ⟶ Cu;      </u>   0.3419

Ni + Cu²⁺ ⟶ Cu + Ni²⁺; 0.599

This is the spontaneous direction.

Cu²⁺ is reduced to Cu.

3. Electrochemical cell

                                        <u>E°/V</u>

Ni²⁺ + 2e⁻ ⟶ Ni;           -0.257

<u>Cu ⟶ Cu²⁺ + 2e⁻;        </u> <u>-0.3419</u>

Cu + Ni²⁺ ⟶ Ni + Cu²⁺; -0.599

This is the non-spontaneous direction.

Ni²⁺ is reduced to Ni in the electrolytic cell.

8 0
3 years ago
The density of mercury is 13.6 g/mL what is the density in lbs/L ( 1 lb hint =0.454 kg )
Vesna [10]

Answer:

30.0\frac{lb}{L}

Explanation:

Hello,

In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:

=13.6\frac{g}{mL} *\frac{1lb}{454g} *\frac{1000mL}{1L} \\\\=30.0\frac{lb}{L}

Regards.

6 0
4 years ago
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