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Archy [21]
2 years ago
15

A chemist dissolves 249.mg of pure perchloric acid in enough water to make up 380.mL of solution. Calculate the pH of the soluti

on. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Alla [95]2 years ago
4 0

Answer:

pH = 2.18

Explanation:

Perchloric acid (HClO₄) is a strong acid. This means that in an aqueous solution it completely dissociates into H⁺ and ClO₄⁻ species.

First we <u>convert 249 mg HClO₄ into moles</u>, using its <em>molecular weight</em>:

  • 249 mg HClO₄ ÷ 100.46 mg/mmol = 2.49 mmol HClO₄

<em>Because it is a strong acid</em>, 2.49 mmol HClO₄ = 2.49 mmol H⁺

We <u>calculate the molar concentration of H⁺</u>:

  • 2.49 mmol H⁺ / 380 mL = 6.52x10⁻³ M

Finally we <u>calculate the pH of the solution</u>:

  • pH = -log[H⁺] = -log(6.52x10⁻³)
  • pH = 2.18

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andriy [413]

Answer:

3 bonds are needed.

Explanation:

The electrons that are involved in chemical bonding are those in the outer shell of the highest energy level of the atom. The electron configuration of nitrogen (N) is 1s²2s²2p³. That means thy at each nitrogen atom has 5 valence electrons: 2 electrons in the 2s orbital and 3 electrons in the 2p orbital. To fullfil the octet, each nitrogen atom needs 3 electrons. So, they can share each other 3 electrons to form 3 simple bonds. Therefore, the nitrogen molecule (N₂) has 3 bonds involving 6 bonding electrons or a triple bond.

8 0
3 years ago
A gas at STP occupies 22.4 L if the temperature is changed to 260 K and the pressures changed it to 0.50 ATM what will the new v
asambeis [7]

Answer:

The new volume will be 42, 7 L.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.

P1xV1/T1 =P2xV2/T2

1 atmx 22,4 L/273K = 0,5atmx V2/260K

V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L

3 0
3 years ago
How many mL of 2M stock solution would I use to prepare 0.500 L of 0.1 M NaCl?
nordsb [41]

Answer:

25 mL

Explanation:

Step 1: Given data

  • Concentration of the concentrated solution (C₁): 2 M
  • Volume of the concentrated solution (V₁): ?
  • Concentration of the diluted solution (C₂): 0.1 M
  • Volume of the diluted solution (V₂): 0.500 L

Step 2: Calculate the volume of the concentrated NaCl solution

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.1 M × 0.500 L / 2 M

V₁ = 0.025 L = 25 mL

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...a metal atom will *lose* electrons to form a *positive* cation and a nonmetal atom will *accept* electrons to form an *negative* anion.
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3 years ago
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