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Ahat [919]
2 years ago
7

The literature value for the heat produced when Mg reacts with H+ under constant pressure conditions is - 110 kcal/mol. Calculat

e the accuracy obtained from this experiment
Chemistry
1 answer:
Naddika [18.5K]2 years ago
6 0

The accuracy obtained from the experiment can be calculated using the formula:

  • Percent accuracy = (obtained value - literature value/literature value) × 100%.

<h3>What is heat of reaction?</h3>

The heat of reaction is the amount of heat evolved or absorbed when reactant molecules react to form products.

The heat of reaction determined under standard conditions is known as standard enthalpy of reaction.

The literature value of enthalpy of reaction Mg and H+ is H+ represents the value under nearly perfectly controlled experimental conditions.

However, in imperfect experimental conditions such as in student laboratories, the value obtained may differ slightly from the literature value.

The accuracy obatined from student experiment can be calculated using the formula:

Percent accuracy = (obtained value - literature value/literature value) × 100%.

Learn more about enthalpy of reaction at: brainly.com/question/14291557

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To understand the relation between the strength of an acid or a base and its pKa and pKb values. The degree to which a weak acid
elena-14-01-66 [18.8K]

Answer:

pKa = 3.675

Explanation:

  • pKa = - Log Ka

∴ <em>C</em> X-281 = 0.079 M

∴ pH = 2.40

let X-281 a weak acid ( HA ):

∴ HA ↔ H+ + A-

⇒ Ka = [H+] * [A-] / [HA]

mass balance:

⇒<em> C</em> HA = 0.079 M = [HA] + [A-]

⇒ [HA] = 0.079 - [A-]

charge balance:

⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water

⇒ [H+] = [A-]

∴ pH = - log [H+] = 2.40

⇒ [H+] = 3.981 E-3 M

replacing in Ka:

⇒ Ka = [H+]² / ( 0.079 - [H+] )

⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )

⇒ Ka = 2.113 E-4

⇒ pKa = - Log ( 2.113 E-4 )

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8 0
3 years ago
Question 2.3
SVETLANKA909090 [29]

Answer:

Your answer is :- C [OH-] = 10 x 10-7 mol dm-3

Mark this as Brainliest

5 0
3 years ago
Read 2 more answers
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