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zimovet [89]
3 years ago
6

Anthracite coal d) is the most abundant grade of coal e) is very soft and burns at high temperatures a) causes the most air poll

ution c) is very hard and burns cleanly b) has the highest sulfur content
Chemistry
1 answer:
forsale [732]3 years ago
6 0

Answer: The correct option is C ( is very hard and burns cleanly).

Explanation:

COAL is a form of rock that is made up of mostly carbon amongst other elements which includes sulphur, nitrogen, hydrogen and oxygen. There are different types of coal which include:

--> anthracite ( 90% carbon)

--> bituminous coal ( 70-90% carbon)

--> lignite ( 60- 70% carbon) and

--> peat (60 % carbon).

Anthracite is the type of coal that contains the highest carbon content ( 90% carbon). This makes it very hard and is often a times referred to as HARD COAL. Anthracite is a higher quality coal for domestic and open fire heating. This is because it contains less impurities than other type of coal and thereby making it to BURN CLEANLY avoiding atmospheric pollution.

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A racing car travels 340 miles in 120 minutes. what speed is traveling at?
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Answer:

It is traveling at 170 miles per hour.

Explanation:

divide 340 by 2 because right now it's at 340 miles per two hours.

You get 170 miles per hour.

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3 years ago
Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0
3 years ago
Complete the following chart to fill in the blanks
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Answer:

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