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gogolik [260]
3 years ago
8

What is the empirical formula of a compound composed of 34.5 g potassium (K) and 7.06 g oxygen (O)?

Chemistry
1 answer:
inessss [21]3 years ago
4 0
Atomic mass : 

K = 39.09 a.m.u
O = 15.99 a.m.u

number of moles:

K = 34.5 / 39.09 => 0.882 moles of K
O = 7.06 / 15.99 => 0.441 moles of O

Use the least number of moles obtained to get the formula empirical:

K = 0.882 / 0.441 => 2
O =  0.441 / 0.441 => 1

Therefore  :  K₂O

 hope this helps!
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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:

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<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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Marta_Voda [28]

Answer:

BaI2

Explanation:

Hello, since the electronegativity of Barium and Iodine are 0.89 and 2.66, respectively, the difference is 1.77, so the bond is ionic.

Best regards.

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4 years ago
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Answer:

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