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1 year ago

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

Chemistry

1 answer:

olga nikolaevna [1]1 year 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:

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How much mercury and oxygen could be obtained from 21.7g of mercury (II) oxide

Musya8 [376]

20.06 g of Hg and 1.6 g of O₂

<u>Explanation:</u>

To Find:

Number of Mercury and oxygen that can be obtained from 21.7 g of HgO

First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,

2 HgO (s) → 2Hg(l) + O₂ (g)

21.7 g of HgO = $\frac{21.7 g}{216.59 g / mol}$

= 0.1 mol of HgO.

As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .

So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)

= 20.06 g of Hg

Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂

Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO

7 0

3 years ago

If you needed a 1.5 x 1 0-4 M solution of a compound that has a molar mass of 760 g/mol, what would it concentration be in part

motikmotik

Answer:

114 ppm

Explanation:

Data obtained from the question include:

Conc. of compound in mol/L = 1.5×10¯⁴ mol/L

Molar mass of compound = 760 g/mol

Conc. in ppm =..?

Next, we shall determine the concentration of the compound in grams per litre (g/L) . This is illustrated below:

Conc. in mol/L = conc. in g/L / Molar mass

1.5×10¯⁴ = conc. In g/L / 760

Cross multiply

Conc. in g/L = 1.5×10¯⁴ x 760

Conc. in g/L = 0.114 g/L

Next, we shall convert 0.114 g/L to milligrams per litre (mg/L). This is illustrated below:

1 g/L = 1000 mg/L

Therefore, 0.114 g/L = 0.114 x 1000 = 114 mg/L

Finally, we shall convert 114 mg/L to parts per million (ppm). This is illustrated below:

1 mg/L = 1 ppm

Therefore, 114 mg/L = 114 ppm

From the calculations made above,

1.5×10¯⁴ mol/L Is equivalent to 114 ppm.

6 0

3 years ago

Can someone please help me on the second quick Check!!!!

alexgriva [62]

#1 fossils paragraph 2

8 0

3 years ago

Read 2 more answers

Consider 80.0-g samples of two different compounds consisting of only carbon and oxygen. One of the compounds consists of 21.8 g

goldenfox [79]

Answer:

<u><em>Ratio of the mass carbon that combines with 1.00 g of oxygen in compound 2 to the mass of carbon that combines with 1.00 g of oxygen in compound 1 = 2</em></u>

Explanation:

First, detemine the mass of oxygen in the two samples by difference:

mass of oxygen = mass of sample - mass of carbon

Item Compound 1 Compound 2

Sample 80.0 g 80.0 g

Carbon 21.8 g 34.3 g

Oxygen: 80.0 g - 21.8g = 58.2 g 80.0 g - 34.3 g = 45.7 g

Second, determine the ratios of the masses of carbon that combine with 1.00 g of oxygen:

For each sample, divide the mass of carbon by the mass of oxygen determined above:

Sample Mass of carbon that combines with 1.00 g of oxygen

Compound 1 21.8 g / 58.2 g = 0.375

Compound 2 34.3 g / 45.7 g = 0.751

Third, determine the ratio of the masses of carbon between the two compounds.

Divide the greater number by the smaller number:

Ratio = 0.751 / 0.375 = 2.00 which in whole numbers is 2

6 0

3 years ago

Drag each tile to the correct location on the image.

nikdorinn [45]

Answer:

[He]: 2s² 2p⁵.

[Ne]: 3s².

[Ar]: 4s² 3d¹⁰ 4p².

[Kr]: 5s² 4d¹⁰ 5p⁵.

[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².

Explanation:

Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.

[He]:

He contains 2 electrons fill 1s (1s²).

So, [He] can be written before the electronic configuration of 2s² 2p⁵.

[Ne]:

Ne contains 10 electrons fill (1s² 2s² 2p⁶).

So, [Ne] can be written before the electronic configuration of 3s².

[Ar]:

Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).

So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².

[Kr]:

Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).

So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.

[Xe]:

Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).

So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².

3 0

3 years ago

Read 2 more answers