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svetoff [14.1K]
2 years ago
14

Where does the light that makes the moon visible come from?

Chemistry
2 answers:
Rzqust [24]2 years ago
6 0
The answer is D. the sun.

Explanation: At night, the moon blocks the sun, while absorbing the light of the sun as-well.
nirvana33 [79]2 years ago
3 0
D. The sun is the correct answer
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What is the electron configuration of bromine whose atomic number is 35
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1s22s22p63s23p64s23d104p5 is the answer
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3 years ago
A molecure that contains 3 iddentical polar bonds to the central atom will be
shepuryov [24]

ANSWER:

Nonpolar

sorry if I'm wrong

4 0
3 years ago
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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3 years ago
If it takes 20757 J to heat 96.5 g of a substance from 18.76°C to 78.62 °C, what is the specific heat of the substance?
Free_Kalibri [48]

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3 years ago
Granulation on the surface of the sun is caused by
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Granules on the photosphere of the Sun are caused by convection currents (thermal columns, Bénard cells) of plasma within the Sun's convective zone. The grainy appearance of the solar photosphere is produced by the tops of these convective cells and is called granulation.

I hope this helps!!!! :)

5 0
3 years ago
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