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m_a_m_a [10]
3 years ago
14

Which change occurs when a barium atom loses two electrons?(1) It becomes a negative ion and its radius decreases.(2) It becomes

a negative ion and its radius increases.(3) It becomes a positive ion and its radius decreases.(4) It becomes a positive ion and its radius increases.
Chemistry
2 answers:
OleMash [197]3 years ago
6 0

Answer:

The correct answer is option 3, It becomes a positive ion and its radius decreases

Explanation:

As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.  

Tcecarenko [31]3 years ago
5 0
<h2>Changes Occurs when a Barium Atom - Option 3  </h2>

When a barium atom loses two electrons it becomes a positive ion and its radius decreases. Barium (Ba) has atomic number 56 so it has 2 electrons in first shell of an atom to become stable according to duplet rule. Then other 52 electrons revolve in the shells according to octet rules.

Another 2 electrons are in the outermost shell. To become stable electrons lose to form barium ions (Ba+2). Hence, by losing 2 electrons the outermost shell will be diminished so its radius decreases and by losing electrons it becomes positive ions.

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A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

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n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

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