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Drupady [299]
3 years ago
11

Which statement is true for a solution when its concentration of hydroxide ions becomes equal to the concentration of hydronium

ions?
It becomes more acidic.

It becomes more alkaline.

Its pH becomes equal to 0.

Its pH becomes equal to 7.
Chemistry
2 answers:
bazaltina [42]3 years ago
7 0

Answer:

Option 4th, TRUE . Its pH becomes equal to 7.

Explanation:

[H₃O⁺] = [OH⁻]

pH = pOH → 7

[H₃O⁺] > [OH⁻]  → pH < 7 ⇒ acidic solution

[H₃O⁺] < [OH⁻]  → pH > 7 ⇒ basic solution

pH = 0 does not exist.

Greeley [361]3 years ago
3 0

Answer : The correct statement is, Its pH becomes equal to 7.

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

Mathematically,

pH=-\log [H^+]

  • When the pH less than 7 then the solution is acidic and the concentration of hydrogen ion is greater than hydroxide ion.
  • When the pH more than 7 then the solution is basic and the concentration of hydrogen ion is less than hydroxide ion.
  • When the pH is equal to 7 then the solution is neutral and the concentration of hydrogen ion is equal to the hydroxide ion.

As per question, the concentration of hydroxide ions becomes equal to the concentration of hydronium ions then its pH becomes equal to 7.

Hence, the correct statement is, Its pH becomes equal to 7.

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Which of the following atoms forms an ionic bond with a sulfur atom? (Points : 3)
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- When the difference of electronegativities between the two atoms is greater than 2.0, then the bond is ionic.

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You need to list the electronegativities of the five elements (there are tables with this information)

Element  electronegativity

Cu:    1.9
H:      2.2
Cl      3.16
I:        2.66
S:      2.58

Differences:

Cu / S: 2.58 - 1.9 = 0.68
H / S: 2.58 - 2.2 = 0.38
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I / S: 2.66 -  2.58 = 0.08

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8 0
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A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature
Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

  • v_0=500
  • \alpha =1.12\times 10^{-4}
  • \Delta \epsilon = 25

Now,

Increase in volume will be:

⇒ \Delta V = \alpha\times v_0\times \Delta \epsilon

By putting the given values, we get

           =1.12\times 10^{-4}\times 500\times 25

           =1.12\times 10^{-4}\times 12500

           =1.4  \ mL

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