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2 years ago

Write a chemical equation for Ba(OH )2 (aq) showing how it is an acid or a base according to the Arrhenius definition.

1 answer:

5 0

Balanced chemical reaction: Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
According to the Arrhenius definition barium hydroxide is base.
According this theory acids are substances which produce hydrogen ions in water solution, bases are substances which produce hydroxide ions in solution. Acids and bases when react (neutralisation) produce salt and water.

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Question 1 Of the following which displays a positive energetic process?

TiliK225 [7]

We can describe a positive energetic process as any process which increases the internal energy of the system.

A positive energetic reaction or process is often referred to as being Endothermic. This means that the system which is performing the process absorbs energy. Some examples include:

Boiling an Egg
Roasting food over a fire (the food is the reference system)

etc

Therefore, we can confirm that a positive energetic process is one in which the system in question absorbs energy, thus increasing its internal energy.

Since I could not locate the options online, I have provided a general explanation of the concept coupled with a few examples.

To learn more visit:

brainly.com/question/4345448?referrer=searchResults

5 0

2 years ago

Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of

Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

Explanation:

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0

3 years ago

A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in a

Lera25 [3.4K]

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

8 0

3 years ago