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3 years ago

In which situation is no work considered to be done by a force?

Chemistry

2 answers:

Masja [62]3 years ago

7 0

If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.

If the angle is 45, then there is still some work involved.

The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90.

san4es73 [151]3 years ago

6 0

Answer:

The correct answer is option B.

Explanation:

Work done is defined as product of applied force to an object and displacement that is distance covered by an object when force applied.

$W=Force\times distance\times \cos\theta$

When $\theta =0^o or 180^o$ . positive or negative work will be done.
When $\theta =45^o$ s, some amount wok will be done
When $\theta =90^o$ no work will be done.

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When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.

EleoNora [17]

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

Answer: The % yield of the product is 87.05 %

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

We are given:

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8mol$

For the given chemical reaction:

$CaCO_3\rightarrow CaO+CO_2$

By stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of CaO

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of CaO

We know, molar mass of $CaO$ = 56 g/mol

Putting values in above equation, we get:

$\text{Mass of CaO}=(0.8mol\times 56g/mol)=44.8g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

$\% \text{yield}=\frac{39 g}{44.8g}\times 100\\\\\% \text{yield}=87.05\%$

Hence, the % yield of the product is 87.05 %

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