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4 years ago

Explain why the chemical formulas of ionic compounds are usually the same as their empirical formulas.

Chemistry

1 answer:

Licemer1 [7]4 years ago

8 0

Answer:

All the ionic compounds are simplest combination of the atoms. All the empirical formulas are the simplest way to write the chemical formulas. Thus, the ionic compound is usually in its empirical state.

Explanation:

Ionic compound are the chemical compound which are composed of the ions which are held together by the strong electrostatic forces which is termed as ionic bonding. The ionic compound is overall neutral, but it consists of positively charged ions which are known as cations and also negatively charged ions which are known as anions.

Since all the ionic compounds are simplest combination of the atoms and also, all the empirical formulas are simplest way to write the chemical formulas, the ionic compound is usually in its empirical state.

Thus it can be stated that:

All the ionic compounds are the empirical formulas, but not all the empirical formulas are the ionic compounds.

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A. A student was given an unknown sample of a pure metal to identify. Its mass was 116

forsale [732]

Answer:

Silver

Explanation:

Given parameters:

Mass = 116g

Energy = 511J

Initial temperature = 21.6°C

Final temperature = 40.5°C

Unknown:

Specific heat capacity and the type of metal = ?

Solution:

To solve this problem, we need to find the specific heat capacity of the metal given,

H = m x c x Ф

H is the energy

m is the mass

c is the specific heat capacity

Ф is the change in temperature

511 = 116 x c x (40.5 - 21.6)

511 = 2192.4c

c = 0.233J/g°C

The specific heat capacity of silver is similar to this.

3 0

3 years ago

According to the kinetic molecular theory, what effect do collisions between gas particles in a sample of an ideal gas have?

elena55 [62]

According to the kinetic theory of gases, the collisions among the molecules are perfectly elastic, that is, when the collision of the molecules takes place, they bounce off each other with no gain or loss of energy. No repulsive or attractive forces take part at the time of collisions, so energy is conserved.

3 0

3 years ago

How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Gnoma [55]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

4 0

3 years ago

Please help this is due today which ones are reactive and which arent?

BARSIC [14]

Answer:

I' really don't know I'm sorry

4 0

3 years ago

A student wants to make 5.00% solution of rubidium chloride using 0.37 g of the substance. what mass of water will be needed to

chubhunter [2.5K]

Answer:-

7.03 g

Explanation:-

Let the mass of water needed be M.

So total mass = M + 0.37

Now 5% of this solution has 0.37g of rubidium chlorude.

∴ (M +0.37) x (5/100) = 0.37

M +0.37 = 0.37 x 100/5

M + 0.37 = 7.4

M = 7.4-0.37

=7.03

Thus water to be added is 7.03 gram

7 0

3 years ago