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lidiya [134]
2 years ago
13

Are my answers correct?

Chemistry
1 answer:
olga_2 [115]2 years ago
6 0

Answer: yes

Explanation: its common sense

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Which is the correct name for p2o5? phosphorus dioxide phosphorus pentoxide diphosphorus pentoxide diphosphorus hexaoxide
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Correct option:

The correct name for P_{2}O_{5}  is diphosphorus pentoxide.

Why P_{2}O_{5} is called diphosphorus pentoxide?

P_{2}O_{5} is commonly known as diphosphorus pentoxide.

Phosphorus pentoxide has an intriguing property in that P_{2}O_{5} is actually its empirical formula, whereas P_{4}O_{10} is its actual molecular formula.

However, the name of the chemical was obtained from its empirical formula rather than from its molecular formula. The official name for this substance is diphosphorus pentoxide.

Oxygen-containing binary compounds have "oxide" as their "last name." Phosphorus is the "first name."

We list each atom's numbers below:

The di- and Penta- prefixes are used to indicate the presence of two and five oxygen atoms, respectively, in the molecule.

Learn more about diphosphorus pentoxide here,

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A chemist dissolves 240mg of pure barium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (Th
goblinko [34]

Answer:

pH = 12.22

Explanation:

<em>... To make up 170mL of solution... The temperature is 25°C...</em>

<em />

The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:

Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)

<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>

<em />

To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:

<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>

0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =

2.80x10⁻³ moles of OH⁻

<em>Molarity [OH⁻] and [H⁺]</em>

2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M

As Kw at 25°C is 1x10⁻¹⁴:

Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]

[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M

<em>pH:</em>

pH = -log [H⁺]

pH = -log [6.068x10⁻¹³M]

<h3>pH = 12.22</h3>
8 0
3 years ago
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