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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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3 years ago

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Explanation:

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2 years ago

The smaller the particle size, the the surface area. Therefore, particle size is to surface area.

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4 years ago

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago

Using the following information, what are the values of the exponents in the rate expression, rate = k[A]x[B]y

pashok25 [27]

Answer:

The correct option is;

A) x = 2, y = 1

Explanation:

Here we have

$k[0.05]^x \cdot [0.05]^y = 0.062 \ M/sec$ ....................(1)

$k[0.05]^x \cdot [0.10]^y = 0.123 \ M/sec$ ....................(2)

$k[0.100]^x \cdot [0.100]^y = 0.491 \ M/sec$ ..............(3)

From experiment (1), (2) it is observed that [A] is held constant and [B] is doubled of which the rate is also observed to be doubled, therefore, when you double a reactant which result in the rate being doubled, the order of the reaction with respect to the reactant is order 1, therefore, y = 1.

Similarly between reaction (2) and (3) when the concentration of the reactant B is doubled and A is held constant the rate of the reaction is multiplied by a factor of 4, therefore, the reaction with respect to that reactant is order 2, therefore, x = 2.

Therefore, the correct option is x = 2, y = 1.

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3 years ago