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Natasha_Volkova [10]
3 years ago
7

Which fossil fuel has the least amount of carbon dioxide released in the air each year

Chemistry
2 answers:
Natali5045456 [20]3 years ago
4 0

The correct answer is - natural gas.

The natural gas is the fossil fuel that releases the least amount of carbon dioxide in the atmosphere. It actually releases very little to none. The reason for that is that the natural gas is a fossil fuel which has the methane as its main ingredient. Even when used, it is not causing some serious pollution, instead the pollution comes from not well managed infrastructure and movement from one place to another of the natural gas. The biggest problem, pollution wise, comes from the leaking of the natural gas, when methane is released into the atmosphere.

Anuta_ua [19.1K]3 years ago
4 0
<h2>Answer:</h2>

<u>Natural gas</u><u> produces the least amount of CO2 in the air</u>

<h2>Explanation:</h2>

The amount of CO2 produced when a fuel is burned is a function of the carbon content of the fuel. The heat content, or the amount of energy produced when a fuel is burned, is mainly determined by the carbon (C) and hydrogen (H) content of the fuel. Natural gas emits 50 to 60 percent less carbon dioxide (CO2) when combusted in air as compared to other fossil fuels

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What is the molarity of a solution that has 2.52 grams of NaCO3 dissolved to
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Explanation:

2.52g/ 0.125L= 20.16M

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Taken all together the reactions of the human body are
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Ibuprofen is a common pain reliever and anti-inflammatory. Its formula is C13H18O2. What percent by mass of ibuprofen is the car
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Answer:

Explanation:

%Carbon =  mass of carbon / mas of C13H18O2) X100%

             =      13X12/ (13X12+18+2X16)X100%

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3 years ago
What is the most likely reason a computer model is used for a black hole?
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6 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
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