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Anuta_ua [19.1K]
2 years ago
10

a balloon containing 1,000L of gas at 50. C and 760 mmHg rises to an altitude where the pressure is 380mmHg and the temperature

is 10. C, what is the new volume of the balloon (in L)?
Chemistry
1 answer:
lubasha [3.4K]2 years ago
5 0

Answer:

V' = 1.75 L

Explanation:

glitterfairy9870

6 hours ago

Chemistry

High School

a balloon containing 1,000L of gas at 50. C and 760 mmHg rises to an altitude where the pressure is 380mmHg and the temperature is 10. C, what is the new volume of the balloon (in L)?

PV/T =P'V'/T'

380 XFIRST, ALWAYS ALWAYS, ALWAYS CHANGE TEMP IN C TO TEMP IN K

50C =50+273K =323K

10C= 10+273 =283

now PV/T =P'V'/T')

so

760 X 1/323 = 380 X V'/283

so

V' =760 X1 X283/(323X380)

V' = 1.75 L

L

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6.0 mol Al reacts with 4.0 mol O2 to form Al2O3.
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Answer:

3.0 moles Al₂O₃

Explanation:

We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.

4 Al + 3 O₂ -----> 2 Al₂O₃

6.0 moles Al            2 moles Al₂O₃
----------------------  x  -------------------------  =  3.0 moles Al₂O₃
                                    4 moles Al

4.0 moles O₂           2 moles Al₂O₃
----------------------  x  -------------------------  =  2.7 moles Al₂O₃
                                    3 moles O₂

As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.

However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.

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What is the mass of 6.02 × 1023 molecules of Fe2O3?
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Explanation: solution attached.

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30 POINTS PLEASE HELP ASAP
lina2011 [118]

Hello!

We have the following data:

f (radiation frequency) = 3.0*10^{19}\:Hz

v (speed of light) = 3.0*10^8\:m/s

λ (wavelength) = ? (in m)

Let's find the wavelength, let's see:

f = \dfrac{v}{\lambda}

3.0*10^{19} = \dfrac{3.0*10^8}{\lambda}

\lambda = \dfrac{3.0*10^8}{3.0*10^{19}}

\boxed{\boxed{\lambda = 1*10^{-11}\:m}}\Longleftarrow(wavelenght)\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

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