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AfilCa [17]
2 years ago
10

(a) What is the charge density of an ion, and what two properties of an ion affect it?

Chemistry
1 answer:
olga nikolaevna [1]2 years ago
6 0

Charge density is the ratio of the charge of ions to the volume of ions.

<h3> Factor affecting charge density</h3>

It depends on the following properties of ion as

  • Charge on the ion.
  • Volume of the ions.

The lithium ion, Li⁺, has half of the radius of the potassium ion, K⁺. That is why, Li⁺ have a electric field 4 times stronger than the electric field of K⁺ and thus it have 4 times stronger force of attraction.

All the ions: Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺, have the similar electric charge +1.6×10⁻¹⁹ C. As we know that, volume is directly proportional to the third power of their radius. Therefore, smaller ions have less charge density than bigger one.

We can say that charge density decreases down across period due to decrease in the radius which further reduces the volume of ions.

Charge density increases along group due to increase in the radius which further increases the volume of ions.

Thus, we can say that charge density depends on the charge and volume of the ions.

learn more about charge density:

brainly.com/question/15810178

#SPJ4

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Answer:

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8 0
2 years ago
A sample of argon gas at 55°C is under 845 mm Hg pressure. What will the new temperature be if the pressure is raised to 1050 mm
Maurinko [17]

Answer:

The final temperature at 1050 mmHg is 134.57 ^{\circ}C or 407.57 Kelvin.

Explanation:

Initial temperature = T = 55^{\circ}C = 328 K

Initial pressure = P = 845 mmHg

Assuming final  to be temperature to be T' Kelvin

Final Pressure = P' = 1050 mmHg  

The final temperature is obtained by following relation at constant volume

\displaystyle \frac{P}{P'}=\displaystyle \frac{T}{T'} \\ \displaystyle \frac{845 \textrm{ mmHg}}{1050 \textrm{ mmHg}} = \displaystyle \frac{328 \textrm{ K}}{T'} \\T' = 407.57 \textrm{ Kelvin}

The final temperature is 407.57 K

7 0
3 years ago
Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.
Oduvanchick [21]

Answer:

103.9 g

Explanation:

  • C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

  • 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

  • 1.23 mol C₃H₈ * \frac{3molCO_2}{1molC_3H_8} = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

  • 3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

  • 162.36 g * 64.0/100 = 103.9 g
7 0
3 years ago
Help I’m stuck on this question :
Dimas [21]

Answer:

The first ionization energy is the energy it takes to remove an electron from a neutral atom.

hope it is helpful :)

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7nadin3 [17]
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