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Ronch [10]
3 years ago
15

You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g) You find the reaction to be first

order with respect to HBr and first order with respect to O2. You propose the following mechanism: HBr(g) + O2(g) → HOOBr(g) HOOBr(g) + HBr(g) → 2 HOBr(g) HOBr(g) + HBr(g) → H2O(g) + Br2(g) a. Confirm that the elementary reactions add to give the overall reaction. (Hint: Use Hess Law) b. Based on the experimentally determined rate law, which step is rate determining? c. What are the intermediates in this mechanism? d. If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

Chemistry
1 answer:
Vlada [557]3 years ago
4 0

Answer:

Explanation:

FIND THE SOLUTION IN THE ATTACHMENT

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Answer:

51.79g Li₃P.

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4 0
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<u>Answer:</u> The outermost valence electron enters the p orbital.

<u>Explanation:</u>

Valence electrons are defined as the electrons which are present in outer most orbital of an atom.

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These 6 electrons enter s-orbital and p-orbital but the outermost valence electron will enter the p-orbital.

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He molecular formula mass of this compound is 180 amu . what are the subscripts in the actual molecular formula?
mezya [45]
I can't actually answer this one if the empirical formula is not given. Luckily, I've found a similar problem from another website. The problem is shown in the picture attached. It shows that the empirical formula is CH₂O. Let's calculate the molar mass of the empirical formula.

Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol

Then, let's divide this to the molar mass of the molecular formula.
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Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
<em>Actual molecular formula = C₆H₁₂O₆</em>

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