PLEASE HELP I HAVE 19 MINUTES LEFT I WILL MARK BRAINLIEST
How much more acidic is a pH of 4 as compared to a pH of 6.5?

Answers

Answer 1

Answer:

316.227766

Explanation:

Answer 2
Answer 3.16 hope it helps

Related Questions

Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.

Answers

"0.0340" mol of CH₄ are in sealed flask.

Methane (CH₄)

Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.

According to the question,

Volume, V = 800 mL or, 0.800 L

Temperature, T = 22°C or, 295

Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm

As we know the relation,

The gram of moles will be will be:

→ n = [tex]\frac{PV}{RT}[/tex]

By substituting the values, we get

     = [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]

     = [tex]\frac{0.824}{242.077}[/tex]

     = 0.0340

Thus the response above is correct.

Find out more information about methane here:

https://brainly.com/question/24690118

The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts

Answers

Answer:

E: the point where the acid and base have been added in proper stoichiometric amounts

Explanation:

Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.

The Equivalence point occurs before the endpoint.

Thus, option E is correct.

Calculate the volume of solvent present in a 55.5%
by volume of 10.5 mL alcohol solution.

Answers

Answer:

I dont know

Explanation:

good luck

A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution

Answers

Answer:

Following are the solution to the given question:

Explanation:

Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.

explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity

Answers

The answer is… Ions are highly charged species that can readily conduct an electrical current. But hydrogen chloride the gas does not exist as ions since there is no solvent medium for them to dissolve into. Hope this helps!

In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?

Answers

I think you forgot to add a picture?

Answer:

Refraction

Explanation:

write half-reactions that show how H2O2 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations occurred in your testing.

Answers

Answer:

H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .

Explanation:

When

H2O2 acts as an oxidizing agent

H2O2 + 2e- 2H+--->   2H2O

Reducing agent

H2O2 --> O2 + 2e + 2H+

H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .

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