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Tanya [424]
3 years ago
9

Calculate the hydronium ion concentration in an aqueous solution with a ph of 9.85 at 25°c.

Chemistry
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer:

1.41 × 10⁻¹⁰ M

Explanation:

We have a solution with a pH of 9.85 at 25 °C. We can calculate the concentration of H⁺ using the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH

[H⁺] = antilog -9.85

[H⁺] = 1.41 × 10⁻¹⁰ M

H⁺ is usually associated with water molecules forming hydronium ions.

H⁺ + H₂O → H₃O⁺

Then, the concentration of H₃O⁺ ions is 1.41 × 10⁻¹⁰ M.

nordsb [41]3 years ago
8 0

Answer:

The hydronium ion concentration for a solution with pH 9.85 = 1.41 *10^-10M

Explanation:

Step 1: data given

pH = 9.85

Temperature = 25.0 °C

the hydronium ion concentration = [H3O+] = [H+]

Step 2: Calculate the hydronium ion concentration

[H3O+] = [H+]

pH = -log[H+]

9.85 = -log[H+]

[H+] = 10^-9.85

[H+] = 1.41 *10^-10M = [H3O+]

The hydronium ion concentration for a solution with pH 9.85 = 1.41 *10^-10M

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Dafna11 [192]

Answer:

5=C, every action has an equal or opposite reaction,

6=B, since it has less air drag and more force exerted on it

7= You're correct

5 0
3 years ago
URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it h
Ede4ka [16]
<h3><u>Answer;</u></h3>

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

<h3><u>Explanation</u>;</h3>

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen  = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = <u>C₂H₃O</u>

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

  n = 7

Therefore;

Molecular formula = (C2H3O)7

                             <u> = C₁₄H₂₁O₇</u>

6 0
3 years ago
Balance the following redox equation using the smallest integers possible and select the correct coefficient for the hydrogen su
goldfiish [28.3K]

Answer:

Coefficient of HSO_{3}^{-} is 5

Explanation:

Oxidation: MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)

  • Balance O and H in acidic medium: MnO_{4}^{-}(aq)+8H^{+}(aq)\rightarrow Mn^{2+}(aq)+4H_{2}O(l)
  • Balance charge : MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l) .............(1)

Reduction: HSO_{3}^{-}\rightarrow SO_{4}^{2-}

  • Balance O and H in acidic medium : HSO_{3}^{-}(aq)+H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq)
  • Balance charge : HSO_{3}^{-}(aq)+H_{2}O(l)-2e^{-}\rightarrow SO_{4}^{2-}(aq)+3H^{+}(aq) ..............(2)

[2\times eq(1)]+[5\times eq(2)] -

Balanced equation: 2MnO_{4}^{-}(aq)+5HSO_{3}^{-}(aq)+H^{+}(aq)\rightarrow 2Mn^{2+}(aq)+5SO_{4}^{2-}(aq)+3H_{2}O(l)

Coefficient of HSO_{3}^{-} is 5

8 0
3 years ago
Water (H2O) is a liquid at room temperature. Ammonia (NH3) is a gas at room temperature. Both are polar covalent compounds. Whic
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the answer is: A. Water

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I say Answer choice A because it is a straight line and it hasn't gone under all those reactions yet. 
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