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grandymaker [24]
3 years ago
15

What is the pH of a solution with an [H+] of (a) 5.4 x 10-10, (b) 4.3 x 10-5, (c) 5.4 x 10-7?

Chemistry
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

a. 9.2

b. 4.4

c. 6.3

Explanation:

In order to calculate the pH of each solution, we will use the definition of pH.

pH = -log [H⁺]

(a) [H⁺] = 5.4 × 10⁻¹⁰ M

pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2

Since pH > 7, the solution is basic.

(b) [H⁺] = 4.3 × 10⁻⁵ M

pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4

Since pH < 7, the solution is acid.

(c) [H⁺] = 5.4 × 10⁻⁷ M

pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3

Since pH < 7, the solution is acid.

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Answer:

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Explanation:

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(Please correct me if I'm wrong)

5 0
2 years ago
Fulgurites are the products of the melting that occurs when lightning strikes the earth. Microscopic examination of a sand fulgu
BlackZzzverrR [31]

Answer:

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Explanation:

In order to determine the empirical formula, we have to follow a series of steps.

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Step 2: Divide each percentage by the atomic mass of the element

Fe: 46.01/55.85 = 0.8238

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Step 3: Divide all the numbers by the smallest one

Fe: 0.8238/0.8238 = 1

Si: 1.922/0.8238 = 2.33

Step 4: Multiply by numbers that make the coefficients whole.

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The empirical formula is Fe₃Si₇.

5 0
3 years ago
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ale4655 [162]

Answer:

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3. The formula of the compound form by X and Y is given as: XY

Explanation:

For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:

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To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:

Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below

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The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

The formula of the compound form by X and Y is given below :

X^2+ + Y^2- —> XY

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4 0
3 years ago
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Your answer is  $20,480.

Hope this Helped!

<h3>Also here is the explanation: </h3><h3>https://www.symbolab.com/solver/equation-calculator/x%3D32000%5Cleft(.8%5Cright)%5E%7B2%7D</h3>
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___AlBr3 + ___K -> ___KBr + ___ Al

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