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PilotLPTM [1.2K]
2 years ago
6

1. What is the pH of a solution with a [H3O+] concentration of 4.32 x 10-2 M?

Chemistry
1 answer:
mel-nik [20]2 years ago
5 0

1. The pH of the solution is 1.36

2. The pH of the solution is 10.47

3. the concentration of [H₃O⁺] in the solution is 6.3×10⁻¹⁴ M

<h3>What is pH ? </h3>

This is simply a measure of the acidity / alkalinity of a solution.

The pH measures the hydrogen ion (hydronium ion) concentration while the poH measures the hydroxide ion concentration. It is expressed mathematically as

pH = –Log H₃O⁺

<h3>1. How to determine the pH</h3>
  • [H₃O⁺] = 4.32×10⁻² M
  • pH = ?

pH = –Log H₃O⁺

pH = –Log 4.32×10⁻²

pH = 1.36

<h3>2. How to determine the pH</h3>
  • [H₃O⁺] = 3.4×10⁻¹¹ M
  • pH = ?

pH = –Log H₃O⁺

pH = –Log 3.4×10⁻¹¹

pH = 10.47

<h3>3. How to determine the concentration of [H₃O⁺]</h3>
  • pH = 13.2
  • [H₃O⁺] = ?

pH = –Log H₃O⁺

13.2 = –Log H₃O⁺

Multiply through by -1

-13.2 = Log H₃O⁺

Take the anti-log of -13.2

H₃O⁺ = anti-log (-13.2)

H₃O⁺ = 6.3×10⁻¹⁴ M

Learn more about pH:

brainly.com/question/3709867

#SPJ1

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How many grams are in 1.76 x 10^23 atoms of iodine
Mariana [72]

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

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