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4 years ago

EXPERIMENT: DETERMINING VOLUME

Chemistry

1 answer:

Tanzania [10]4 years ago

7 0

Explanation:

(Ques- A) Why does the first method for determining volume work only for a regular-shaped object?

(Ans- A) Because the method requires precise dimensions of objects for result, which is not possible for irregular shaped objects.

(Ques - B) Will the second method for determining volume work for any object or just an odd-shaped one? Why?

(Ans-B) It will work for both regular and irregular shaped objects since both displace equal volumes of water.

(Ques - C) Is one method of measurement more accurate than the other? Why or why not?

(Ans-C) Both are pretty accurate, with some experimental errors which may creep in accidentally.

(Ques- D) Would the displacement method of measurement work for a cube of sugar? What about a cork? Why?

(Ans - D) No, the method would not work because sugar being soluble, will dissolve in water.

No, the method would not work because sugar being soluble, will dissolve in water. Cork is less dense than water so floats on it, with only part of it submerged in water, resulting in displacement of less volume of water than actual volume of Cork.

(Ques-E) What did you find out from this investigation? Be thoughtful in your answer.

(Ans- E) I learnt about determining volume of different objects from this investigation. (Sorry, I know its not a very thoughtful answer)

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Answer : The mass of reactant $H_2$ remain would be, 0.20 grams.

Solution : Given,

Moles of $H_2$ = 0.40 mol

Moles of $O_2$ = 0.15 mol

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $H_2$

So, 0.15 moles of $O_2$ react with $0.15\times 2=0.30$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

The moles of reactant $H_2$ remain = 0.40 - 0.30 = 0.10 mole

Now we have to calculate the mass of reactant $H_2$ remain.

$\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(0.10moles)\times (2g/mole)=0.20g$

Therefore, the mass of reactant $H_2$ remain would be, 0.20 grams.

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Answer:

See Explanation

Explanation:

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