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MaRussiya [10]
3 years ago
10

If you conduct a single displacement reaction and you observe bubbles, what are some possibilities of the composition of those g

as bubbles? pick all that apply
a.ammonia
b.air
c.oxygen gas
d.chlorine gas
e. hydrogen gas
Chemistry
1 answer:
pickupchik [31]3 years ago
5 0

Answers:

    c. oxygen gas

    d. chlorine gas

    e. hydrogen gas


Justification:


Single displacement reactions are those reactions in which the atoms of one element substitute (replace or displace) the atoms of other element in a compound.


The representación of this ind of reactions is:

  • A + BX → AX + B: the atoms of the element A displace the atoms of the element B in the compound BX. As result, the element B stands separated after the reaction.

The conditions for a single displacemente reactions are:

  • One single element react with a compound
  • The reactant single element will form part of a new compound
  • One single element is removed from the compound
  • The one single element removed from the compound stands as a separated product.
  • It is a more active element displaces a less active element from the compound.

In the giiven statement, the element displaced will be as a gaseous product.


Ammonia is not a single element, but the compound with the formula NH₃, so it is not the result of a single displacement reaction, and it is not a possibility (discarded)


Air is not either a single element nor a compound, so it is not the result of a single displacemente reaction, and it is not a possibility (discarded).


Since oxygen is one of the most reactive non-metal elements, oxygen gas is usually the result of a decomposition reaction and not of a single displacement reaction.


Decomposition reactions are those in which a single compound breaks down into two or more simpler substances (elements and/or compounds). Some examples of decompositon reaction in which oxygen gas evolves as a product are:


  • The decomposition of water into hydrogen and oxygen gases.
  • The decomposition of hydrogen peroxide to form oxygen gas and water.
  • The decomposition of solid potassium chlorate to form solid potassum chloride and oxygen gas.

Nevertheless, fluorine is more active than oxygen, and so it can oxidises water according to this single replacement reaction:

  • 2F₂ + 2H₂O → 4HF + O₂↑

As you see, fluorine displaces oxygen, so this is one of the right choices.


Regarding chlorine, being it less active than fluorine (fluorine is the most active halogen and the most active gas) you can predict without doubt that it can be displaced by flouorine. And so, chloringe gas is a right choice too.


The evolution hydrogen gas is a typical example of single displacement reactions and can be seen in this example:

  • 2Li(s) + 2H₂O → 2LiOH + H₂(g)↑

Hence, hydrogen gas is one of the right options.


Conclusion: we have tested the choices one by one and the result is that the possibilities of the composition of those gas bubbles are:

   c.oxygen gas

    d.chlorine gas

    e. hydrogen gas


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<u>Answer:</u> The molecular formula for the compound is C_6H_6

<u>Explanation:</u>

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Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

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Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = \frac{7.68}{7.68}=1

For Hydrogen = \frac{7.74}{7.68}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

  • <u>Calculating the molar mass of the compound:</u>

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{m}{M}RT

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L  (Conversion factor:  1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = 62.3637\text{ L. torr }mol^{-1}K^{-1}

T = temperature of the gas = 100^oC=(100+273)K=373K

Putting values in above equation, we get:

820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

n=\frac{78.10g/mol}{13g/mol}=6

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 6)}H_{(1\times 6)}=C_6H_6

Hence, the molecular formula for the compound is C_6H_6

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