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Gnoma [55]
3 years ago
7

1.Which statement best describes an oxidation-reduction reaction?(1 point)

Chemistry
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

D

Explanation:

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Calculate the number of ammonia molecules in 3.9 g.
disa [49]
•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol

1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:

(3.9 g)/ (17.03g/mol) = 0.22900763mols

Then convert the moles to molecules by multiplying it with Avogadro’s number:

Avogadro’s number: 6.022 x 10^23


0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
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3 years ago
A chemistry student needs 55.0g of carbon tetrachloride for an experiment. by consulting the crc handbook of chemistry and physi
Monica [59]

The volume of a substance is simply the ratio of mass and density. Therefore:

volume = mass / density

 

Calculating for volume of Carbon Tetrachloride that the student has to pour out:

volume = 55.0 g / (1.59 g / cm^3)

<span>volume = 34.60 cm^3</span>

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2 years ago
The plant chemical which absorbs sunlight and helps make cell food is
UkoKoshka [18]

Answer:

chlorophyll

Explanation:

8 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

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Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

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7 0
2 years ago
54,000,000 nanometers equals 0.000000054 megameters.
LiRa [457]
That is true b<span>ecause 1 nanometer = 1.0 × 10^-15 megameters.</span>
3 0
3 years ago
Read 2 more answers
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