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3 years ago

An aqueous solution at 25 °C has a pOH of 12.42. Calculate the pH. Round your answer to 2 decimal places

Chemistry

2 answers:

Helga [31]3 years ago

5 0

Answer:

The pH of a solution that has a pOH of 12.42 is 1.58

Explanation:

Step 1: Data given

At a temperature of 25.0 °C the pOH = 12.42

pH + pOH = 14

Step 2: Calculate the pH

pH + pOH = 14

pH = 14.00- pOH

pH = 14.00 - 12.42

pH = 1.58

The pH of a solution that has a pOH of 12.42 is 1.58

love history [14]3 years ago

3 0

Answer:

pH=1.58

Explanation:

As we know that

pH+pOH=14

Given that pOH=12.42

we get pH=14-pOH

or pH=14-12.42

or pH=1.58

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What mass of natural gas (ch4) must you burn to emit 272 kj of heat?

balandron [24]

CH4 + 2 O2 ---> CO2 + 2 H2O Q = 891,6 kJ / mol CH4

1 mol CH4 = 16 g

16 g ---- 891,6 kJ
x g ----- 272 kJ

x = 272 kJ × 16 g / 891,6 kJ = 4,88 g

You must burn 4,88 g of CH4.

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3 years ago

Which of the following reactions have a positive ΔSrxn? Check all that apply.

PolarNik [594]

Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

2A(g) + B(s) → 3C(g)

2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

2A(g) + B(s) → 3C(g)

1) The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

A(g) + B(g) → C(g)

Again, 2 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.

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3 years ago

What type of reaction is: sulfuric acid + potassium hydroxide -> potassium sulfate + water?

aleksley [76]

Answer:

Double replacement reaction

Explanation:

Now, let us first write the reaction equation properly:

H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O

The above reaction is a neutralization reaction between an acid and a base whose product gives salt and water only at most instances.

From here, we can observe that the species displaces on another in their ionic state. Hydrogen replaces potassium and water is produced. Potassium combines chemically with sulfate ions to give the salt of potassium.

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3 years ago

For alkyl halides used in SN1 and SN2 mechanisms, rank the leaving groups in order of reaction rate. You are currently in a rank

Alex777 [14]

Answer:

Iodide> Bromide > chloride > flouride

Explanation:

During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.

This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).

In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.

Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.

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