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mafiozo [28]
3 years ago
10

How do I Balancing Chemical Equations

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
6 0

Answer:

what?

Explanation:

You might be interested in
Please help
dem82 [27]

Answer:Because binary ionic compounds are confined mainly to group 1 and group 2 elements on the one hand and group VI and VII elements on the other, we find that they consist mainly of ions having an electronic structure which is the same as that of a noble gas. In calcium fluoride, for example, the calcium atom has lost two electrons in order to achieve the electronic structure of argon, and thus has a charge of +2:By contrast, a fluorine atom needs to acquire but one electron in order to achieve a neon structure. The resulting fluoride ion has a charge of –1:The outermost shell of each of these ions has the electron configuration ns2np6, where n is 3 for Ca2+ and 2 for F–. Such an ns2np6 noble-gas electron configuration is encountered quite often. It is called an octet because it contains eight electrons. In a crystal of calcium fluoride, the Ca2+ and F– ions are packed together in the lattice shown below. Careful study of the diagram shows that each F– ion is surrounded by four Ca2+ ions, while each Ca2+ ion has eight F– ions as nearest neighbors.

Thus there must be twice as many F– ions as Ca2+ ions in the entire crystal lattice. Only a small portion of the lattice is shown, but if it were extended indefinitely in all directions, you could verify the ratio of two F– for every Ca2+. This ratio makes sense if you consider that two F– ions (each with a –1 charge) are needed to balance the +2 charge of each Ca2+ ion, making the net charge on the crystal zero. The formula for calcium fluoride is thus CaF2.Figure 6.10.1

6.10.

1

: A portion of the ionic crystal lattice of fluorite, calcium fluoride. (a) Ca2+ ions (color) and F– ions (gray) are shown full size. “Exploded” view shows that each F– surrounded by four Ca2+ ions, while each Ca2+ ion is surrounded by eight F– ions. The ratio of Ca2+ ions to F– ions is thus 4:8 or 1:2, and the formula is CaF2. (Computer-generated). (Copyright © 1976 by W. G. Davies and J. W. Moore.)

Newcomers to chemistry often have difficulty in deciding what the formula of an ionic compound will be. A convenient method for doing this is to regard the compound as being formed from its atoms and to use Lewis diagrams. The octet rule can then be applied. Each atom must lose or gain electrons in order to achieve an octet. Furthermore, all electrons lost by one kind of atom must be gained by the other.

An exception to the octet rule occurs in the case of the three ions having the He 1s2 structure, that is, H–, Li+ and Be2+. In these cases two rather than eight electrons are needed in the outermost shell to comply with the rule.

Example 6.10.1

6.10.

1

: Ionic Formula

Find the formula of the ionic compound formed from O and Al.

Solution

We first write down Lewis diagrams for each atom involved:

alt

We now see that each O atom needs 2 electrons to make up an octet, while each Al atom has 3 electrons to donate. In order that the same number of electrons would be donated as accepted, we need 2 Al atoms (2 × 3e– donated) and 3 O atoms (3 × 2e– accepted). The whole process is then

alt

The resultant oxide consists of aluminum ions, Al3+, and oxide ions, O2–, in the ratio of 2:3. The formula is Al2O3.Figure 6.10.1

6.10.

1

: A portion of the ionic crystal lattice of fluorite, calcium fluoride. (a) Ca2+ ions (color) and F– ions (gray) are shown full size. “Exploded” view shows that each F– surrounded by four Ca2+ ions, while each Ca2+ ion is surrounded by eight F– ions. The ratio of Ca2+ ions to F– ions is thus 4:8 or 1:2, and the formula is CaF2. (Computer-generated). (Copyright © 1976 by W. G. Davies and J. W. Moore.)

Newcomers to chemistry often have difficulty in deciding what the formula of an ionic compound will be. A convenient method for doing this is to regard the compound as being formed from its atoms and to use Lewis diagrams. The octet rule can then be applied. Each atom must lose or gain electrons in order to achieve an octet. Furthermore, all electrons lost by one kind of atom must be gained by the other.

An exception to the octet rule occurs in the case of the three ions having the He 1s2 structure, that is, H–, Li+ and Be2+. In these cases two rather than eight electrons are needed in the outermost shell to comply with the rule.

Example 6.10.1

6.10.

1

: Ionic Formula

Find the formula of the ionic compound formed from O and Al.

Solution

We first write down Lewis diagrams for each atom involved:

alt

We now see that each O atom needs 2 electrons to make up an octet, while each Al atom has 3 electrons to donate. In order that the same number of electrons would be donated as accepted, we need 2 Al atoms (2 × 3e– donated) and 3 O atoms (3 × 2e– accepted). The whole process is then

alt

The resultant oxide consists of aluminum ions, Al3+, and oxide ions, O2–, in the ratio of 2:3. The formula is Al2O3.

Explanation:

3 0
2 years ago
My high school chemistry is learning this and I don't understand it. "Heat is measured by the following equation:
tangare [24]

Answer:

you didn't ask a question so here is your explanation.

Explanation:

Q = mc∆T. Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/kg∙K) ∆ is a symbol meaning "the change in"

5 0
3 years ago
How many moles of sodium ions are present in 0.25 kg of a 2.0 m solution of sodium chloride in water
LenKa [72]

0.50 is the answer on edge

8 0
3 years ago
Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3
Pachacha [2.7K]

<u>Answer:</u> The volume of barium chlorate is 195.65 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL

Hence, the volume of barium chlorate is 195.65 mL

6 0
3 years ago
The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.
Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions {\text{IO}_{3}}^{-}.

How many moles of iodine \text{I}_2 will be produced?

{\text{IO}_{3}}^{-} converts to \text{I}_2 in the first reaction. The coefficient in front of \text{I}_2 is three times the coefficient in front of {\text{IO}_{3}}^{-}. In other words, each mole of {\text{IO}_{3}}^{-} will produce three moles of \text{I}_2. 0.0020 moles of {\text{IO}_{3}}^{-} will convert to 0.0060 moles of \text{I}_2.

How many moles of thiosulfate ions {\text{S}_2\text{O}_3}^{2-} are required?

\text{I}_2 reacts with {\text{S}_2\text{O}_3}^{2-} in the second reaction. The coefficient in front of \text{I}_2 is twice the coefficient in front of {\text{S}_2\text{O}_3}^{2-}. How many moles of {\text{S}_2\text{O}_3}^{2-} does each mole of \text{I}_2 consume? Two. 0.0060 moles of \text{I}_2 will be produced. As a result, 2 \times 0.0060 = 0.0120 moles of {\text{S}_2\text{O}_3}^{2-} will be needed.

6 0
3 years ago
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