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3 years ago

How can these tiles be considered a model of an ionic solid?

Chemistry

1 answer:

coldgirl [10]3 years ago

5 0

Answer:

The Black square are considered as the lattice points (Atoms)

$\blacksquare$ = lattice point

Since there is regular arrangement of atoms , therefore it is considered as the ionic solid.

Explanation:

Consider the following points :

Ionic Solids : Those solids which are formed by continuous arrangement of cations(positive charge) and anions(negative charge) bounded by strong force of attraction are called ionic solids.

When These cations and anions arrange regularly in space, they form Crystal Lattice.

Crystal Lattice : A regular 3-D arrangement if the points in space .

Here the whole figure is Crystal Lattice.

Lattice Points : Each Point in the lattice is called lattice point.

One lattice point represent one Particle , it can be the atom , the molecue or the ions .

In ionic solid these lattice points are ions

$\blacksquare$ = Lattice Points = ions

These lattice points are joined by a straight line to bring out the Geometry of the Lattice.

So In the given figure :

$\blacksquare$ = Ions

The figure as whole = Crystal lattice

And also known as Ionic Solid

Hence these ties can be considered as the model of ionic solids.

You might be interested in

How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)

nikdorinn [45]

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

Na = 22.99 g

H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

$200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}$

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200. * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.

8 0

3 years ago

1. Your friend has a balloon that can hold a volume of 5 L. If she added a 3 mol of CO2 gas at a pressure of 0.90 atm, what is t

masha68 [24]

Answer:

I don't understand your question

6 0

3 years ago

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g CO2 and 0.6551 g

Ede4ka [16]

Answer:

C2H4O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.6004/44 = 0.0364

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0364= 0.4368g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.6551/18 = 0.0364 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0364= 0.0728

The mass of hydrogen is thus 0.0728* 1 = 0.0728g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.8009 - 0.0728 - 0.4368 = 0.2913 mole

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.2913/16 = 0.0182 mole

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.0182

H = 0.0728/0.0182 = 4

C= 0.0364/0.0182 = 2

O= 0.0182/0.0182= 1

The empirical formula is thus C2H4O

3 0