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Shtirlitz [24]
3 years ago
7

A solution of hydrofluoric acid, HF(aq), is at equilibrium. How would the system change if more HF were added to the solution?

Chemistry
2 answers:
pav-90 [236]3 years ago
8 0

More H+ and F would be produced.

Option D.

<h3><u>Explanation:</u></h3>

Hydrofluoric acid is one of the strong acid, but not as strong as hydrogen chloride. This is because the fluorine is much small atom which strongly attracts the Electron of hydrogen, but also exhibits covalent bond.

So in solution, hydrogen fluoride partially breaks into hydrogen ions and fluoride ions.

So when in solution, they exhibit an equilibrium between the HF and hydrogen ions and fluoride ions. So when more amounts of hydrogen fluoride is added to solution, the concentration of hydrogen fluoride increases in solution. This shifts the reaction to the reactant side, and hence more dissociation of the hydrogen and fluoride ions will occur and this will produce more hydrogen ions and fluoride ions.

zzz [600]3 years ago
6 0

Answer:

D. More H+ and F would be produced.

Explanation:

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2 years ago
How do you convert grams of zinc to moles of zinc
leva [86]
Take mass of zinc divided by the relative molecular mass - 30 for zinc ( from periodic table)
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3 years ago
What is the correct order of the steps in the scientific<br> method?
Gnom [1K]

Answer:

Make an observation.

Ask a question.

Form a hypothesis, or testable explanation.

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7 0
3 years ago
rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0
3 years ago
Read 2 more answers
How many significant figures are in the measurement 0.03050
faltersainse [42]
There are 4 significant figures(3050)
8 0
3 years ago
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