Let's write the reaction first:
<span>2 Au + 2 BrF</span>₃<span> + 2 KF → 2 KAuF</span>₄<span> + Br</span>₂<span>
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To know which of the substances is the oxidizing agent, calculate the oxidation numbers of each element before and after the reaction. If the element's oxidation number decreases, then that is the oxidation agent. Let's calculate Br.
BrF₃: x + 3(-1) = 0; x = +3
Br₂: 0 (for any elemental form, oxidation number is 0)
The oxidation number is reduced from +3 to 0. <em>Thus, BrF₃ is the oxidizing agent.</em>
Answer:
sugar crystals with stirring at 15°C
Explanation:
just took the quiz
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
The endotherdic energy raises that its breaks down the bond of the solvent which makes the solute-slovent interaction weaker
