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user100 [1]
3 years ago
8

What are the environmental impacts of coal

Chemistry
2 answers:
mixas84 [53]3 years ago
8 0

Answer:

Look down below

Explanation:

Environmmetal impacts that coal can cause is Sulfur dioxide (SO2), which contributes to acid rain and respiratory illness.

AND.....

Nitrogen oxides (NOx), which contribute to smog and respiratory illnesses. Particulates, which contribute to smog, haze, and respiratory illnesses and lung disease

Scilla [17]3 years ago
3 0

Answer:

Several principal emissions result from coal combustion: Sulfur dioxide (SO2), which contributes to acid rain and respiratory illnesses.

Explanation:

<h3><em>hehe.</em></h3>

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Give the volume indicated. Be sure to include all significant figures.
miskamm [114]

Answer: -

71

Explanation: -

From the diagram, we see that volume goes from 70 to 75 in 5 markings.

Each marking is for 1.

It is the most accuracy possible as there is no smaller marking.

Significant figures expresses the required amount of accuracy.

Thus the volume indicated from the diagram is 71 .

8 0
3 years ago
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The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
3 years ago
Which macromolecules are polymers made of nucleotides
VMariaS [17]
Nucleic Acids are polymers made of nucleotideds 
8 0
3 years ago
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A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
3 years ago
How do you convert 34.6 kilograms to micrograms?<br> Also the Steps that got you that answer
saw5 [17]
40.28 sorry if it is wrong:) but I tried
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4 years ago
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