Answer:
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Explanation:
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In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:

Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.
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When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
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The mole ratio of iron(III) oxide to produced iron is 1:2.
Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles
Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles
Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams
More on stoichiometric calculations can be found here; brainly.com/question/27287858
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Answer:
138.96kJ is the maximum electrical work
Explanation:
The maximum electrical work that can be obtained from a cell is obtained from the equation:
W = -nFE
<em>Where W is work in Joules,</em>
<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>
Zn(s) → Zn²⁺(aq) + 2e⁻
F is faraday constant = 96500Coulombs/mol
E is cell potential = 0.72V
Replacing:
W = -2mol*96500Coulombs/mol*0.72V
W = - 138960J =
<h3>138.96kJ is the maximum electrical work</h3>
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I'm sure that to calculate the freezing point depression <span>subtract</span> solution's freezing point and the freezing point of it's pure solvent. According to the formula.