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3 years ago

Give an example of how environmental science might involve geology and chemistry.

1 answer:

8 0

One of the examples is radiation and chemistry of water. Environmental science requires the capacity to integrate data from the greater part of the significant fields of science, and in addition from arithmetic.
Geology is vital on the grounds that huge scale arrives forms make geology. The presence of mountains and valleys influences how much daylight and precipitation achieve the ground, how breezy an area is, the manner by which precipitation keeps running off, and numerous different variables that figure out what plants and creatures will have the capacity to occupy a district.

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A meteor is moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement desc

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2 years ago

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Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

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

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

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2 years ago

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Answer:

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3 years ago

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