Answer:
pH = 12.22
Explanation:
<em>... To make up 170mL of solution... The temperature is 25°C...</em>
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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>
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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>
Explanation:
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Volume is an extensive property of a marker ink and color is an intensive property of a marker ink.
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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