What can you do to help reduce Acid Rain? List 5 ways. Help me yall!! Imao

Chemistry

1 answer:

chubhunter [2.5K]2 years ago

6 0

Answer:

- to produce energy without using fossil fuels instead using renewable sources solar and wind power.

- reduce the amount of sulfur dioxide released from coal-burning power plants. One option is to use coal that contains less sulfur.

- install equipment called scrubbers, which remove the sulfur dioxide from gases leaving the smokestack.

- While one car alone does not produce much pollution, all the cars on the road added together create lots of pollution. Therefore, car manufacturers are required to reduce the amount of nitrogen oxides and other pollutants released by new cars. One type of technology used in cars is called a catalytic converter. This piece of equipment has been used for over 20 years to reduce the amount of nitrogen oxides released by cars.

- recycling is one way that is often use less energy than producing a new products from raw materials so the use of energy that obtain mainly from fossil fuel will be reduced and that reduced the emissions of toxic gases.

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A chemical formula written above or below the yield sign indicates:

sweet-ann [11.9K]

I believe the correct answer from the choices listed above is option B. A chemical formula written above or below the yield sign indicates <span>that the substance is used as a catalyst. I am certain with this answer. Hope this helps. Have a nice day.</span>

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3 years ago

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A hot air balloon is filled to 1250 m3 at 27 C. At what temperature will the balloon be filled to 1600 m3 if the pressure remain

SSSSS [86.1K]

Answer:

384.2 K

Explanation:

First we convert 27 °C to K:

27 °C + 273.16 = 300.16 K

With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:

T₁V₂=T₂V₁

Where in this case:

T₁ = 300.16 K

V₂ = 1600 m³

T₂ = ?

V₁ = 1250 m³

We input the data:

300.16 K * 1600 m³ = T₂ * 1250 m³

And solve for T₂:

T₂ = 384.2 K

7 0

2 years ago

A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th

Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.

The mass percent of lithium hydroxide can be calculated with the following equation:

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$ (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$ (2)

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$