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qwelly [4]
3 years ago
10

Write the chemical equation for hbro reacting with water. what are the products? do you expect to have any hbro molecules left i

n solution when the reaction is completed/reaches equilibrium?
Chemistry
2 answers:
pashok25 [27]3 years ago
8 0

Answer:

1. Reaction:

HBrO+H_2O-->H_3O^{+}+BrO^{-}

2. Yes, the solution will have HBrO molecules.

Explanation:

Hello,

In this case, the undergoing chemical reaction involving hypobromous acid and water yields both hypobromite and hydronium ions as shown below:

HBrO+H_2O-->H_3O^{+}+BrO^{-}

Now, since hypobromous acid's pKa is about 2x10⁻⁹, such value indicates that it is not completely dissociated in this reaction, therefore, there will be an amount of hypobromous acid molecules even when the equilibrium is attained.

Best regards.

Vinil7 [7]3 years ago
6 0
1) HOBr stands for hypobromous acid. On reacting with water, products formed are OBr- and H3O+. Following reaction occurs during this process.

<span>                                     HOBr + H2O </span>⇄<span> OBr- + H3O+

2) HOBr is a weak acid and have a lower value of dissociation constant (Ka ~ </span><span>2.3 X 10^–9). Hence, </span><span> large number of undissociated HOBr molecules are left in solution, when the reaction is completed/reaches equilibrium.</span>


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Answer:

B) K2X

Explanation:

In an uncharged compound, the total oxidation state must be zero. The oxidation state of the calcium is +2, thus we get the following formula, where x is the oxidation state of the polyatomic ion X:

x + 2 = 0 \\ x = 0 - 2 \\ x =  - 2

Also, it is known that potassium has an oxidation state of +1. Since the new compound also has a total oxidation state equal to zero, we get the following equation, where k is the number of K atoms:

k \times 1 + ( - 2) = 0 \\ k - 2 = 0 \\ k =  - ( - 2) \\ k = 2

That's how it is found that the compumd consists of 2 K+ ions and one X ion.

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PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro
podryga [215]

Answer :

Part 1 : Balanced reaction, 3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

Part 2 : The theoretical yield of NH_3 gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of N_2 = 475 g

Molar mass of N_2 = 28 g/mole

Molar mass of NH_3 = 17 g/mole

Experimental yield of NH_3 = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

<u>Answer for Part (2) :</u>

First we have to calculate the moles of N_2.

\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles

From the given reaction, we conclude that

1 moles of N_2 gas react to give 2 moles of NH_3 gas

16.96 moles of N_2 gas react to give \frac{2}{1}\times 16.96=33.92 moles of NH_3 gas

Now we have to calculate the mass of NH_3 gas.

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g

Therefore, the theoretical yield of NH_3 gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100

\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%

Therefore, the % yield of ammonia is 90.03 %

3 0
3 years ago
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