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Alexus [3.1K]
2 years ago
10

A helium balloon with an internal pressure of 1atm and a volume of 4.20 L at 18.0°C is released. What volume will the balloon oc

cupy at an altitude where the pressure is 0.500 atm and temperature is -10.0°C?
Chemistry
1 answer:
kari74 [83]2 years ago
3 0

Answer:

7.59 L

Explanation:

- Use combined gas law formula and rearrange.

- Change C to K

- Hope that helped! Please let me know if you need further explanation.

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Based in the rules of John Dalton, what is a possible hypothesis that could be tested in the Atomic Theory lab?
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8 0
3 years ago
Read 2 more answers
Could volume by displacement be used to determine the volume of a lump of rock salt? Explain would be nice
Readme [11.4K]

Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:

1).  Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water.  So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease.  If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.

So you would have to choose some other fluid besides water ... one in
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be.  You'd have to shop around and find one.

2).  Whatever fluid you did choose, it would also have to be less dense
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5 0
3 years ago
The difference in the energy exerted by an 8.0 earthquake compared to a 6.0 earthquake?
hammer [34]
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4 0
3 years ago
How many hydrogen atoms are found in the<br> hydrate (NH4)3PO4.5H2O?
igomit [66]

Answer:

22

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3x4+  5x2

3 0
2 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
2 years ago
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