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Marta_Voda [28]
3 years ago
8

A student claims that if she wanted to make a solution quickly, she should use small pellets instead of powder along with heatin

g and stirring. Do you agree or disagree with the student's claim? Answer using CER format. PLS HELP I’m a little confused and don’t know what to put.
Chemistry
1 answer:
torisob [31]3 years ago
4 0

Answer:

<em>I disagree with the student's claim</em>

<em></em>

Explanation:

<em>From basic chemistry, the rate of chemical reaction increases with an increased surface area.</em>

<em>This simply means that a powdered substance will react faster than a small pellet of the same substance.</em>

Let us break this down further. The pellet has a smaller exposed surface per grain to the solvent, here is how. Imagine the powdered substance bunched up together to make the small pellet, grains at the center of the pellet is blocked by other grains around it, and will take time before it comes in contact with the solvent. For the powdered grains, each grain has a surface independently exposed to the solvent and can easily come in contact with the solvent, taking less time to react with the solvent.

Technically, a powdered substance has more exposed surface area and logically will take a shorter time for the reaction to be complete.

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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
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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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