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Slav-nsk [51]
3 years ago
6

Both renewable and nonrenewable resources are used within our society. How do the uses of nonrenewable resources compare to the

uses of renewable resources?
a.
Nonrenewable resources have more applications than renewable resources.
b.
Certain types of renewable energy can be used for as many applications as certain types of nonrenewable resources.
c.
There are more sources of nonrenewable energy than renewable energy.
d.
Nonrenewable resources have fewer applications than renewable resources.
Chemistry
2 answers:
madreJ [45]3 years ago
7 0

Renewable sources are which can be gained again and again from nature

for example : sun energy or wind energy

non renewable: these once consumed cannot be regained

example : fossil fuels

Now both can be used for many kinds of applications however the resources so we can say that Certain types of renewable energy can be used for as many applications as certain types of nonrenewable resources

Mila [183]3 years ago
4 0

Renewable energy has the potential to have all the same applications as non-renewable. But we currently don't have the resources and potential.

So I would say the answer would be B, but this question is somewhat confusing.

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What is the mass in grams of 9.76 x 10 12 atoms of naturally occurring sodium?
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Now,

          if     6.02 × 10²³ atoms are found in 1 mole ofsodium
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⇒     x  =    (9.76 × 10¹² )  ÷  (6.02 × 10²³)

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4 0
3 years ago
What is the molar concentration of chloride ions in a
Yanka [14]

Answer:

The concentration of chloride ions in the final solution is 3 M.

Explanation:

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number of moles = concentration in molarity * volume

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For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:

number of moles of Cl⁻ = 2 * number of moles of  CaCl₂

number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻

The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.

Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:

Concentration = number of moles of Cl⁻ / volume

Concentration = 150.2 mol / 50.1 l = 3.0 M

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