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?
1 answer:
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Answer:
9.64g/mL
Explanation:
Given parameters:
Mass of the metal = 106g
Volume of cylinder = 50mL
Volume difference = 31mL - 20mL = 11mL
Unknown:
Density of the metal = ?
Solution:
To find the density of the metal, we use;
Density =
Density = = 9.64g/mL
Answer:
A: 1,2-dimethylcyclopropane
Explanation:
The possible cyclic structure with formula C₅H₁₀ are shown in the image.
A is a cyclic compound. On monochlorination, A yields 3 products.
To have 3 products on monochlorination, there should be three different carbon atoms.
Considering structure 1, all carbons have same nature, thus only one product will be formed and thus not a structure of A.
Considering structure 2, there are two different carbon atoms, thus two different structure are formed and thus not a structure of A.
Considering structures 3 and 4 , there are four different carbon atoms, thus four products will be formed and either of them are not a structure of A.
Considering structure 5, there are three different carbon atoms, thus three different structure are formed and thus the A is structure 5.
