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Fofino [41]
3 years ago
12

Discuss the environmental problems caused by the combustion of fossil fuels(a few lines only)

Chemistry
1 answer:
sp2606 [1]3 years ago
5 0
Burning fossil fuels release a lot of carbon dioxide which is a greenhouse gas which contributes to global warming. Also burning of fossil fuels causes a lot of air pollution. You can also describe in detail the effect of carbon dioxide on global warming in general
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A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?
zhannawk [14.2K]
POH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁹ ] 

pOH = 9

Answer C

hope this helps!
3 0
3 years ago
1. Explain the two processes that keep atmospheric oxygen and carbon dioxide at stable levels?
lesya [120]

Cellular Respiration and Photosynthesis. Photosynthesis is the process of when plants use sunlight to make foods from carbon dioxide and water, to later on make oxygen. Cellular respiration is the process through which cells convert sugars into energy.

4 0
3 years ago
Qué tipo de reacción es el siguiente? CH4 + 02<br><br> CO2 + H2O
Tatiana [17]

Answer:

Combustion reaction

Explanation:

Let's consider the following balanced equation.

CH₄ + 2 O₂  ⇒ CO₂ + 2 H₂O

This reaction is known as a combustion reaction, in which a compound reacts with oxygen to form a compound of carbon and water.

  • If the product is carbon dioxide, the combustion is complete.
  • If the product is carbon monoxide or carbon, the combustion is incomplete.
7 0
3 years ago
Read 2 more answers
One characteristic of a scientific theory is that ____.
faust18 [17]
It really base on facts ,hypotheses,evidences,concepts,
8 0
3 years ago
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How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?
Korolek [52]

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead =  177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g       (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead =\frac{177.7}{207.2}\times 1310=1123kJ=1123000J

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

7 0
3 years ago
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