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3 years ago

9

What is the meaning of weight?

2 answers:

zmey [24]3 years ago

8 0

<span>1. (It depends on th example)a body's relative mass or the quantity of matter contained by it, giving rise to a downward force; the heaviness of a person or thing."he was at least 175 pounds in weight"<span>synonyms:<span>heaviness, mass, load, burden, pressure, force; <span>More</span></span></span><span />2.a heavy object, especially one being lifted or carried.<span /><span /></span>verb<span>1.hold (something) down by placing a heavy object on top of it."a mug half filled with coffee weighted down a stack of papers"2.attach importance or value to."speaking, reading, and writing should be weighted equally in the assessment"</span>

Lostsunrise [7]3 years ago

4 0

Weight is properly defined as how much gravity acts upon an object.
I hope this helps!

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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

<u>Answer:</u> The % yield of the product is 87.05 %

<u>Explanation:</u>

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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