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Delvig [45]
3 years ago
6

Gaseous compound Q contains only xenon and oxygen. When 0.100 gg of Q is placed in a 50.0 mLmL steel vessel at 0 ∘C∘C, the press

ure is 0.229 atm
Chemistry
1 answer:
asambeis [7]3 years ago
5 0

The question is incomplete, here is the complete question:

Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.229 atm. What is the likely formula of the compound?

A. XeO

B. XeO_4

C. Xe_2O_2  

D. Xe_2O_3

E. Xe_3O_2

<u>Answer:</u> The chemical formula of the compound is XeO_4

<u>Explanation:</u>

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

PV = nRT

Or,

PV=\frac{w}{M}RT

where,

P = Pressure of the gas = 0.229 atm

V = Volume of the gas  = 50.0 mL = 0.050 L     (Conversion factor:  1 L = 1000 mL)

w = Weight of the gas = 0.100 g

M = Molar mass of gas  = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gas = 0^oC=273K

Putting value in above equation, we get:

0.229\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.229\times 0.050}=195.4g/mol\approx 195g/mol

The compound having mass as 195 g/mol is XeO_4

Hence, the chemical formula of the compound is XeO_4

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According to libretexts the answer would be B. decreases.

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Explanation:

To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;

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When Be loses two electrons it becomes isoelectronic with He;

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Read that From picture and tell me the answer please​
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Answer:

1g Hydrogen

Explanation:

<h3><u>Getting</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>equation</u><u>:</u></h3>

Calcium in water reacts vigorously to give a cloudy white <em>Precipitate</em><em> </em>(compound) called Calcium hydroxide alongwith the evolution of Hydrogen gas.

\boxed{ \mathsf{Ca + H_2O \rightarrow Ca(OH)_2 + H_2}}

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<h3><u>Balancing</u><u> </u><u>the</u><u> </u><u>equation</u><u>:</u><u> </u></h3>

This reaction is not in it's balanced form! The number of atoms of Hydrogen on the left is 2 while that on the right is 4,I.e.,they're not equal.

Adding a 2 in front of H2O solves the problem by making the number of atoms of each element on both the sides equal.

\mathsf{Ca +2 H_2O \rightarrow Ca(OH)_2 + H_2}

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Observations</u><u>:</u></h3>

Looking into the equation more carefully, we see:

<u><em>1</em><em> </em></u><em>atom</em><em> </em><em>of</em><em> </em><em>Calcium</em><em> </em><em>reacts</em><em> </em><em>with</em><em> </em><em><u>2</u></em><em> </em><em>molecules</em><em> </em><em>of</em><em> </em><em>water</em><em> </em><em>to</em><em> </em><em>give</em><em> </em><u><em>1</em><em> </em></u><em>molecule</em><em> </em><em>of</em><em> </em><em>Calcium</em><em> </em><em>Hydroxide</em><em> </em><em>alongwith</em><em> </em><em><u>1</u></em><em> </em><em>molecule</em><em> </em><em>of</em><em> </em><em>Hydrogen</em><em> </em><em>gas</em><em>.</em>

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Gram</u><u> </u><u>atomic</u><u> </u><u>and</u><u> </u><u>molecular</u><u> </u><u>masses</u><u> </u></h3>

<u>Mass</u><u> </u><u>of</u><u> </u><u>one</u><u> </u><u>atom</u><u> </u><u>of</u><u> </u><u>Calcium</u><u> </u>= it's gram atomic mass

= 40 g

<u>Mass of one "molecule" of Hydrogen</u>

= it's Gram molecular mass

= gram mass of one atom × number of atoms in one molecule

= 1 × 2

= 2 g

So,

according to our observation:

One atoms of Calcium gives one molecule of Hydrogen <em>(during the particular reaction)</em>

=><u> 40g of Calcium gives = 2g of Hydrogen</u>

•°• 1 g of Calcium gives = \frac{2}{40}

= \frac{1}{20} g Hydrogen

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Answer</u><u>:</u></h3>

We're provided with 20g of Calcium,

=> 20g of Calcium gives = 20 × \frac{1}{20} g H2

<u>= 1 g H2</u>

_______________

Hope this helps!

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