Gold (III) nitrate in an aqueous solution is hydrolyzed with formation of gold (III) metahydroxide.
Au(NO₃)₃ → Au³⁺(aq) + 3NO₃⁻(aq)
Au³⁺ + H₂O ⇄ AuOH²⁺ + H⁺
AuOH²⁺ + H₂O ⇄ Au(OH)₂⁺ + H⁺
Au(OH)₂⁺ + H₂O → AuOOH·H₂O(s) + H⁺
Au(NO₃)₃(aq) + 2H₂O(l) = AuOOH(s) + 3HNO₃(aq)
Answer:
See figure 1
Explanation:
In this question, we have to start with the <u>protonation of the double bond</u>. In carvone we have two double bonds, so, we have to decide first which one would be protonated.
The problem states that the <u>terminal alkene</u> is the one that would is protonated. Therefore, we have to do the <u>protonation</u> in the double bond at the bottom to produce the <u>carbocation number 1</u>. Then, a hydride shift takes place to produce the <u>carbocation number 2</u>. A continuation, an <u>elimination reaction</u> takes place to produce the <u>conjugated diene</u>. Then the diene is protonated at the <u>carbonyl group</u> and with an elimination reaction of an hydrogen in the <u>alpha carbon</u> we can obtain <u>carvacol. </u>
Answer:
6.25%
Explanation:
Given data:
Half life of lutetium-117 = 6.75 days
Percentage remaining after 27 days = ?
Solution;
Number of half lives = Time elapsed / half life
Number of half lives = 27 days / 6.75 days
Number of half lives = 4
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At 3rd half life = 25%/2 = 12.5%
At 4th half life = 12.5%/2 = 6.25%
Answer
Rapid Combustion
Explanation:
An explosion is a combustion and a quick reaction, That makes it rapid
Answer:
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Explanation: