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Tomtit [17]
3 years ago
13

What are the steps needed when solving problem like this?

Chemistry
2 answers:
a_sh-v [17]3 years ago
5 0

Answer: The number of fluorine atoms that combine with one uranium atom are 6.

Explanation:

We are given an unknown gas of uranium and fluorine atom and when it is decomposed, it produces 2.09 parts of uranium atoms for every 1 part of fluorine atom.

Relative mass of Uranium = 238 g

Relative mass of Fluorine = 19 g

Number of atoms of uranium present in a given relative mass = \frac{2.09}{238}=0.0087atoms

Number of atoms of fluorine present in given relative mass = \frac{1}{19}=0.052atoms

For 0.0087 atoms of uranium are produced when 0.05 atoms of fluorine are produced.

So, 1 atom of uranium are produced when = \frac{0.052}{0.0087}\times 1=5.97\approx 6 atoms of Fluorine atoms.

Hence, 6 atoms of fluorine atoms are produced when 1 uranium atom is produced.

emmainna [20.7K]3 years ago
3 0
For the answer to the question above,the gas can be decomposed to yield 2.09 parts by mass of uranium for every 1 part by mass of fluorine.If the relative mass of a uranium atom is 238 and the relative mass of a fluorine atom is 19,  the number of fluorine atoms that are combined with one uranium atom. is 6
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Liquefied Natural Gas

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3 years ago
A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

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Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

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Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

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Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

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