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laila [671]
1 year ago
6

You should be able to see small bubbles forming where the bromine solution falls into the flask. if the humidity is high enough

you may notice fumes rising out of the mixture what are the fumes?
Chemistry
1 answer:
allochka39001 [22]1 year ago
6 0

If the humidity is high enough we may notice fumes rising out of the mixture containing Br solution. The fumes produced are HBr fumes.

<h3>What is hydrogen bromide? </h3>

Hydrogen bromide (HBr) is a corrosive and toxic gas. It can be fatal if it is inhaled. It is very corrosive to our respiratory system, to our eyes, and when it is in contact with skin. When it is in mist form or in a liquid form it will cause burns to all tissue if it falls upon.

<h3>Why fumes are produced? </h3>

Hydrogen Bromide is a highly irritating, corrosive, nonflammable gas having a pungent odor at room temperature. It will fume in air which have high humidity because formation of hydrobromic acid takes place; it is the diatomic molecule.

Under ordinary conditions hydrogen bromide is a gas but it can be converted into liquid. Hydrobromic acid is then formed after hydrogen bromide is dissolved in water.

Thus we concluded that the fumes produced when bromine solution is in contact with moist air is HBr fumes.

learn more about HBr:

brainly.com/question/2712747

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Given:

<span>BaO2(s) + 2 HCl(aq) => H2O2(aq) + BaCl2(aq) 
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Solution: To determine the amount of hydrogen peroxide that would be produced, we use need to determine which is the limiting reactant and use the initial amount of this reactant for the calculations. 

0.0279 g HCl / mL ( 25.1 mL ) ( 1 mol HCl / 36.46 g HCl) ( 1 mol BaO2 / 2 mol HCl ) = 0.0096 mol BaO2 needed
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Therefore, the limiting reactant would be barium peroxide since it is consumed completely first in the reaction.

1.49 g BaO2 ( 1 mol BaO2 / 169.3 g BaO2 ) ( 1 mol H2O2 / 1 mol BaO2 ) ( 34.02 g H2O2 / 1 mol H2O2 ) = 0.299 g H2O2
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How many ions are in sodium bicarbonate
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Answer:

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Explanation:

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In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe
AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

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\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq).

The first test tube used to contain \text{NH}_4\text{OH}. \text{NH}_4\text{OH} is a weak base that dissociates partially in water.

\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

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\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

The indicator equilibrium will shift to the right to produce more {\text{H}_3\text{O}}^{+} ions along with the colored indicator ions. The solution will show a pink color.

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The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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