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Nikolay [14]
3 years ago
14

A Rolaids tablet contains calcium for neutralizing stomach acid. if a Rolaids tablet neutralizes 24.65ml of 0.547m hydrochloric

acid, how many milligrams of calcium carbonate are in a Rolaids tablet?​
Chemistry
1 answer:
Lelu [443]3 years ago
4 0

There are 674 mg of calcium carbonate in a Rolaids tablet

<h3>Further explanation</h3>

Reaction

CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ +H₂O

V HCl = 24.65 ml

M HCl = 0.547

mol HCl :

\tt 24.65\times 0.547=13.48~mlmol

ratio mol CaCO₃ : HCl = 1 : 2, so mol CaCO₃ :

\tt \dfrac{1}{2}\times 13.48=6.74~mlmol=6.74\times 10^{-3}~mol

mass of CaCO₃ (MW=100~g/mol) :

\tt 6.74\times 10^{-3}\times 100=0.674~g=674~mg

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The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
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The activation energy for this reaction = 23 kJ/mol.

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Using the expression,

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Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

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The conversion of T( °C) to T(K) is shown below:

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So,  

T = (280 + 273.15) K = 553.15 K  

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The conversion of T( °C) to T(K) is shown below:

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So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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