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Elden [556K]
3 years ago
13

You are standing on the rim of a canyon, you drop a rock and in 7.0 seconds hear the sound pof it hitting the bottom, how deep i

s the canyon ? what assumptionsdid you make? examine how each assump[tion affects the answer. Does it lead to a larger or smaller depth than the calculated depth
Chemistry
1 answer:
sp2606 [1]3 years ago
8 0

This should help you with your question:

Gravitational pull = 9.8 m/s^2

1/2 x 9.8 x 7^2 = 240.1 m

The rock has fallen 240.1 m.

I cannot help you with the rest, but this should give you a headstart.

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According to the following reaction, what volume in milliliters of 0.244M KCl(aq) solution is required to react exactly with 50.
Elodia [21]

Answer:

There is 86.1 mL of KCl needed

Explanation:

<u>Step 1:</u> Data given

Molarity of KCl = 0.244 M

Volume of a 0.210 M Pb(NO3)2 = 50.0 mL = 0.05 L

<u>Step 2:</u> The balanced equation

2KCl(aq) + Pb(NO3)2(aq) -> PbCl2(s) + 2KNO3(aq)

<u>Step 3:</u> Calculate moles Pb(NO3)2

moles Pb(NO3)2 =molarity * volume

moles Pb(NO3)2 = 0.210 M * 0.05 L

moles Pb(NO3)2 = 0.0105 moles

<u>Step 4:</u> Calculate moles of KCl

For 1 mole of Pb(NO3)2 we need 2 moles of KCl

moles KCl required = 0.0105 * 2 =0.0210  moles

<u>Step 5:</u> Calculate volume of KCl

Volume = moles KCl / molarity KCl

V = 0.0210 moles / 0.244 M=0.0861 L = 86.1 mL

There is 86.1 mL of KCl needed

6 0
2 years ago
Solve using significant figures <br> 8.647 + 45.969
Murrr4er [49]

Answer:

54.616

Explanation:

8.647 + 45.969

or rewrite for easier look:

45.969 +

 8.647 =

54.616

Hope this helped :3

 

5 0
3 years ago
Read 2 more answers
How many atoms are in 1.6g C? Answer in units of atoms. <br><br> (no answer choices were given)
maksim [4K]

Answer:

6.022  ×10(index 23) / 7.5 = 0.8293 ×10(index 23)

Explanation:

molar mass of C =  12gmol

therefore in 12g of C there is one mole or an amount of 6.022 ×10(index 23)

∴12g/6.02210(index 23) ×1.6g

3 0
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Westkost [7]
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6 0
2 years ago
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balu736 [363]

Answer:

6FCRXCTV

Explanation:

  1. ED5RF6GT7HY8JU9KI0LO-;PJH7YNG6GD
7 0
2 years ago
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