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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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) determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting 27.9 g ko2 with 29.0 l of c
Airida [17]
The Balanced Chemical Equation is as follow;

                         4 KO₂  +  2 CO₂    →    2 K₂CO₃  +  3 O₂

First find out the Limiting Reagent,
According to equation,

         284 g (4 moles) KO₂ reacted with  =  44.8 L (2 moles) of CO₂
So,
                  27.9 g of KO₂ will react with  =  X  L of CO₂


Solving for X,
                          X  =  (44.8 L × 27.9 g) ÷ 284 g

                          X  =  4.40 L of CO₂

Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,

According to eq.

         284 g (4 moles) KO₂ formed  =  138.2 g of K₂CO₃
So,
         27.9 g of KO₂ will form  =  X g of K₂CO₃

Solving for X,
                        X  =  (138.2 g × 27.9 g) ÷ 284 g

                        X  =  13.57 g of K₂CO₃

So, 13.57 g of K₂CO₃ formed is the theoretical yield.

%age Yield  =  13.57 / 21.8 × 100

%age Yield  =  62.24 %
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