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r-ruslan [8.4K]
3 years ago
7

Charlie is balancing an equation. She has identified the atoms and counted the number of each in the reactants and products. Wha

t would Charlie adjust to make the number of atoms in the reactants the same as the number of atoms in the products?
Chemistry
2 answers:
Alecsey [184]3 years ago
8 0
The answer to your question is Coefficients. 

*Hope this helps you! Feel free to message me if you have any further questions. Have a nice day! :)*
VladimirAG [237]3 years ago
7 0

<u>Answer:</u> Charlie must add coefficients in front of the compounds that are written in the equation.

<u>Explanation:</u>

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. In a chemical reaction, total number of atoms on the reactant side must be equal to the total number of atoms on the product side.

<u>For Example:</u> The formation of water follows the given chemical equation:

2H_2+O_2\rightarrow 2H_2O

The coefficients 2 is added in front of hydrogen and water molecule, so that the total atoms becomes balanced.

Hence, Charlie must add coefficients in front of the compounds that are written in the equation.

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Read the scenario and then answer the question.
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Answer: Option 1.

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6 0
3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
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