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Talja [164]
2 years ago
6

A gas occupies 900.0 mL at a temperature of 300K. What is the volume at 132.0 °C?

Chemistry
1 answer:
lozanna [386]2 years ago
8 0
V1/T1=V2/T2
V2=(V1)(T2)/T1
Plug in values given (for the temp you can either turn 300K to 27°C or turn 132°C into kelvin
V2= 4400 mL= 4.4L
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A mole of oxygen and a mole of hydrogen (at STP) have all of the following in common EXCEPT
Drupady [299]

Answer:

Root mean squared velocity is different.

Explanation:

Hello!

In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:

v_{rms}=\sqrt{\frac{3RT}{MM} }

Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.

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8 0
2 years ago
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vazorg [7]

Answer:

Write a balanced chemical reaction:

N2 + 3H2 ==> 2NH3

Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3.  So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:

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Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed

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8 0
2 years ago
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Tresset [83]

Answer:

0.166 m/s

Explanation:

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6 0
2 years ago
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leva [86]
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2 years ago
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Ulleksa [173]

Answer:

ok

Explanation:

Which of the following is NOT a possible clue that a chemical change has

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A change of color

O A change of state

Production of gas

Formation of a precipitate

3 0
3 years ago
Read 2 more answers
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