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Contact [7]
3 years ago
15

If the hydrogen ion concentration, [H+], in a solution is 4.59 x 10-6 M, what is [OH-]?

Chemistry
1 answer:
atroni [7]3 years ago
6 0

Answer:

2.18x10^-9 M

Explanation:

From the question given,

Hydrogen ion concentration, [H+] = 4.59x10^-6 M

Hydroxide ion, [OH-] =?

The hydroxide ion concentration, [OH-] in the solution can be obtained as follow:

[H+] x [OH-] = 1x10^-14

4.59x10^-6 x [OH-] = 1x10^-14

Divide both side by 4.59x10^-6

[OH-] = 1x10^-14 / 4.59x10^-6

[OH-] = 2.18x10^-9 M

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<u>Explanation:</u>

  • Oxy-fuel welding is a process that utilizes fuel gases and oxygen to weld metals. Oxyfuel gas or flame refers to a group of welding processes that utilize the flame delivered by the blending of fuel gas and oxygen as the source of heat.
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Answer:

See below.

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The ""size"" of an atom is sometimes defined by the radius of a sphere that contains 90% of the charge density of the electrons.
vfiekz [6]

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3/2a

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The complete step by step answer is found in the attachment

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PLEASE HELP!! I REALLY NEED HELP
Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

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<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

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2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

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<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

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Answer:

The answer is in the explanation.

Explanation:

The dissociation of a weak acid consist in the following equilibrium:

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A strong acid (HY) dissociates completely in water, thus:

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As the strong acid produces H⁺, in the equilibrium, the reaction shifts to the left -The undissociated form-, reducing the production of H⁺, allowing ignore the dissociation of the weak acid when calculating the pH.

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