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ella [17]
3 years ago
8

Is there a difference between a homogeneous mixture of hydrogen and oxygen in a 2:1 ratio and a sample of water vapor?

Chemistry
1 answer:
Feliz [49]3 years ago
5 0

Answer:

Yes

Explanation:

There is a difference between the homogeneous mixture of the hydrogen and the oxygen in a 2:1 ratio and the sample of the water vapor.

In the homogeneous mixture of the hydrogen and the oxygen which are present in the ratio, 2:1 , the elements are not chemically combined. They are explosive also as both shows their specific properties. They can be separated by physical means (Condensation, diffusion).

On the other hand, in water vapor, the two elements are chemically bonded in a specific mixture which cannot be separated via physical means. Water has its unique properties and they can be separated by chemical means only.

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4 years ago
How many moles are in 3.01 x 10 atoms of magnesium
Fiesta28 [93]
There are 3.01 moles. round to 3 if it asks for a whole number 
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You are playing a game “will it float?” In this game, you are given a large, square can of tuna. If you know the density of wate
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3 years ago
Calculate the number of moles of C2H6 in 6.29×1023 molecules of C2H6.
Kipish [7]

1.05moles

Explanation:

Given parameters:

Number of molecules of C₂H₆ = 6.29 x 10²³molecules

Unknown:

Number of moles = ?

Solution:

The mole is the amount of substances that contains Avogadro's number of particles i.e 6.02 x 10²³

 To find the number of moles:

  number of moles = \frac{number of particles}{Avogadro's number}

  number of moles = \frac{6.29 x 10^{23} }{6.02 x 10^{23} }

  number of moles = 1.05moles

Learn more:

moles brainly.com/question/1841136

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7 0
3 years ago
An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C
ICE Princess25 [194]

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=\frac{1680}{336}

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C

7 0
3 years ago
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