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german
3 years ago
12

Which of the following is a true statement about acids and bases?

Chemistry
2 answers:
maks197457 [2]3 years ago
5 0

Answer: The correct answer is 'a base gives off hydroxide ions'.

Explanation:

  • Bases have pH more than seven.
  • They give hydroxide ions,OH^-.

BOH(aq)\rightarrow B^+(aq)+OH^-(aq)

  • Acids have pH lower than seven.
  • They give hydronium ions,H^+.

HA(aq)\rightarrow H^+(aq)+A^-(aq)

juin [17]3 years ago
3 0

Answer: A base gives off hydroxide ions.

Explanation: According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions (OH^-) when dissolved in water and an acid is defined as a substance which donates hydronium ions (H_3O^+) in water.

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

Acids have pH values ranging from 1 to 6.9, neutral solutions have pH of 7 and bases have pH ranging from 7.1 to 14.

A substance that increases the hydroxide ion concentration of a solution is an Arrhenius base.

NaOH\rightarrow Na^++OH^-

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1.51986 atm

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7 0
3 years ago
A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, w
Fed [463]

Answer:

Density = 8.92 g/cm³

Explanation:

Density shows the relation of the mass and volume of a determined object. We have this deffinition:

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First of all, we calcualte the volume of the block of copper metal, with the data given:

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Now we replace at the density formula:

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7 0
3 years ago
What is a nucleus, chromosome, chlooplasts,cell membrane, cytoplasm?
stiks02 [169]

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8 0
3 years ago
A salt is formed when an acld and _____ react.<br>A.Water<br>B.another acid<br>C.base.​
Dovator [93]
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4 0
3 years ago
One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the
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Answer:

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Explanation:

Given:

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Computation:

ΔS surrounding (entropy change of the reservoir) = - ΔH / T

ΔS surrounding (entropy change of the reservoir) = - 150 / 150

ΔS surrounding (entropy change of the reservoir) = -1 J/K

3 0
4 years ago
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