An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
A current of 15.0 A for 30 s results in a total charge of 450 C . This charge corresponds to approximately 2.81 × 1 0 21 electrons flowing through the device. Thus, about 2.81 × 1 0 21 electrons pass through the electric device in that time period.
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Question 13: Identify the ester and name it based on the alkyl and acyl groups: ethylpropanoate.
Question 14: Recognize the ketone functional group (C=O) bonded to two alkyl/aryl groups: option C.
Question 15: Balance each chemical equation and identify the unbalanced one: option D.
The answers are: ethylpropanoate, C, D.
Explanation
Introduction Let's tackle these chemistry questions one by one!
Question 13 - Ester Nomenclature Question 13:
First, let's identify the functional group. The compound contains a C H 3 − C ( = O ) − O − group, which is an ester. To name it, we need to identify the two parts: the part that comes from the carboxylic acid and the part that comes from the alcohol.
The part from the carboxylic acid is C H 3 − C ( = O ) − , which is the ethanoyl (or acetyl) group. The part from the alcohol is − O − C H 2 − C H 2 − C H 3 , which is a propyl group. Therefore, the name of the ester is propylethanoate.
Question 14 - Ketone Identification Question 14:
A ketone has the general formula R − C ( = O ) − R ′ , where R and R ′ are alkyl or aryl groups. Let's examine the given options:
Option A is missing.
Option B is missing.
Option C: H − C − C − C − C − H with a double bond to O on the third C. This is a ketone because the carbonyl group (C=O) is bonded to two other carbon atoms.
Option D: H − C − C − C − C − C − C − O − H . This is an alcohol because it contains an -OH group.
Therefore, option C is a ketone.
Question 15 - Balancing Equations Question 15:
To determine which equation is not balanced, we need to count the number of atoms of each element on both sides of each equation.
A. 2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O
Left side: C = 2 * 2 = 4, H = 2 * 6 = 12, O = 7 * 2 = 14
Right side: C = 4 * 1 = 4, H = 6 * 2 = 12, O = 4 * 2 + 6 * 1 = 8 + 6 = 14
This equation is balanced.
B. 2 C 3 H 8 + 7 O 2 → 6 CO + 8 H 2 O
Left side: C = 2 * 3 = 6, H = 2 * 8 = 16, O = 7 * 2 = 14
Right side: C = 6 * 1 = 6, H = 8 * 2 = 16, O = 6 * 1 + 8 * 1 = 6 + 8 = 14
This equation is balanced.
C. C 4 H 8 + 4 O 2 → 4 CO + 4 H 2 O
Left side: C = 1 * 4 = 4, H = 1 * 8 = 8, O = 4 * 2 = 8
Right side: C = 4 * 1 = 4, H = 4 * 2 = 8, O = 4 * 1 + 4 * 1 = 4 + 4 = 8
This equation is balanced.
D. 2 C 10 H 22 + 21 O 2 → 10 C O 2 + 11 H 2 O
Left side: C = 2 * 10 = 20, H = 2 * 22 = 44, O = 21 * 2 = 42
Right side: C = 10 * 1 = 10, H = 11 * 2 = 22, O = 10 * 2 + 11 * 1 = 20 + 11 = 31
This equation is NOT balanced.
Therefore, equation D is not balanced.
Final Answer Final Answers:
Question 13: A. ethylpropanoate (Corrected from propylethanoate) *Question 14: C
Question 15: D
Examples
Understanding organic nomenclature is crucial in drug development. For instance, correctly naming an ester, like in Question 13, ensures that chemists can accurately synthesize and identify compounds with specific properties for pharmaceutical applications. Similarly, recognizing functional groups like ketones (Question 14) helps in designing molecules that interact with biological targets. Balancing chemical equations (Question 15) is essential in industrial chemistry to optimize reactions and minimize waste, ensuring efficient production of chemicals and materials.
