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KengaRu [80]
3 years ago
6

Balance the following: C₂ + B → B₂C₂ Ag₄ + SO₂ → AgO + S₂

Chemistry
1 answer:
jarptica [38.1K]3 years ago
3 0

Answer:

The balanced reactions are:

C₂ + 2B → B₂C₂

Ag₄ + 2SO₂ → 4AgO + S₂

Explanation:

The Law of Conservation of Matter postulates that "the mass is not created or destroyed, only transformed." This means that the reagents interact with each other and form new products with physical and chemical properties different from those of the reagents because the atoms of the substances are ordered differently. But the amount of matter or mass before and after a transformation (chemical reaction) is always the same, that is, the quantities of the masses involved in a given reaction must be constant at all times, not changing in their proportions when the reaction ends. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

In other words, the law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts .

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.  

You have:

<em>C₂ + B → B₂C₂</em>

Left side: 2 C and 1 B.

Right side: 2 C and 2 B.

Carbon C is balanced since you have the same amount of the element on each side of the reaction, but boron B is not balanced. So, balancing the reaction:

<u><em>C₂ + 2B → B₂C₂</em></u>

Left side: 2 C and 2 B.

Right side: 2 C and 2 B.

Now you have the same amount of each element on each side of the reaction, so the reaction is balanced.

Now you have:

<em>Ag₄ + SO₂ → AgO + S₂</em>

Left side: 4 Ag, 1 S and 2 O.

Right side: 1 Ag, 2 S and 1 O.

Balancing the Ag:

<em>Ag₄ + SO₂ → 4AgO + S₂</em>

Left side: 4 Ag, 1 S and 2 O.

Right side: 4 Ag, 2 S and 4 O.

Balancing the O:

<u><em>Ag₄ + 2SO₂ → 4AgO + S₂</em></u>

Left side: 4 Ag, 2 S and 4 O.

Right side: 4 Ag, 2 S and 4 O.

You have the same amount of each element on each side of the reaction, so the reaction is balanced.

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The first order equation for radioactive decay can be expressed as :

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Decay constant can be calculated using half life . Decay constant and half life can be related as :

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Given : Half life of Strontium -90 = 28.8 years

Plugging value of T_\frac{1}{2} in above formula (equation 2) :

28.8 yrs = \frac{ln 2}{ k }

Multiply both side by k

28.8 yrs * k = \frac{ln 2 }{k} * k

Dividing both side by 28.8 yrs

\frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs}

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Plugging these value in equation (1) as :

ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t

ln (0.0971 ) = -0.0241 yrs ^-^1 * t

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\frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1}

t = 96.68 yrs

Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs

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