Use this condensed chemical structure to complete the table below. OH I CH3- -CH-OH Some facts about the 1,1-ethanediol molecule: number of carbon-carbon single (C – C) bonds: number of carbon-hydrogen single (CH) bonds: number of oxygen-hydrogen single (O – H) bonds: number of lone pairs: X The condensed chemical structure of 1,1-ethanediol 8 000
Pls help asap and make sure your answer is correct. Also can you pls explain how did you get the answer. I am little confused what I am doing wrong there, thank you
Use this condensed chemical structure to complete the table below. OH I -CH-OH CH3- The condensed chemical structure of 1,1-ethanediol Some facts about the 1,1-ethanediol molecule: number of carbon-carbon single (C – C) bonds: number of carbon-hydrogen single (C – H) bonds: number of oxygen-hydrogen single (O – H) bonds: number of lone pairs: X 0
The correct answer and explanation is:
Problem 1: Factors of 1080 divisible by 10 but not by 30
Step 1: Prime Factorization of 1080
1080=23⋅33⋅511080 = 2^3 \cdot 3^3 \cdot 5^1
A factor is divisible by 1010 if it contains at least 212^1 and 515^1. A factor is divisible by 3030 if it contains 212^1, 313^1, and 515^1.
Step 2: Total factors divisible by 10
To count factors divisible by 1010, fix 2≥12 \geq 1 and 5≥15 \geq 1. The exponents of 22, 33, and 55 can vary as follows:
- 2k2^k where k=1,2,3k = 1, 2, 3
- 3m3^m where m=0,1,2,3m = 0, 1, 2, 3
- 5n5^n where n=1n = 1
The total number of such factors: 3⋅4⋅1=123 \cdot 4 \cdot 1 = 12
Step 3: Total factors divisible by 30
Fix 2≥12 \geq 1, 3≥13 \geq 1, and 5≥15 \geq 1. The exponents of 22, 33, and 55 can vary as:
- 2k2^k where k=1,2,3k = 1, 2, 3
- 3m3^m where m=1,2,3m = 1, 2, 3
- 5n5^n where n=1n = 1
The total number of such factors: 3⋅3⋅1=93 \cdot 3 \cdot 1 = 9
Step 4: Factors divisible by 10 but not 30
12−9=312 – 9 = 3
Answer: (b) 3
Problem 2: Factors of 1920 satisfying conditions
Step 1: Prime Factorization of 1920
1920=27⋅31⋅511920 = 2^7 \cdot 3^1 \cdot 5^1
Step 2: Conditions
- Perfect square: All exponents in the prime factorization must be even.
- Multiple of 10: Contains at least 212^1 and 515^1.
- Not divisible by 30: Cannot contain 313^1.
Step 3: Satisfying all conditions
- For a perfect square, 272^7 contributes even powers: 20,22,24,262^0, 2^2, 2^4, 2^6 (4 choices).
- 313^1 cannot be present (00 choice).
- 515^1 contributes 505^0 or 525^2 (11 choice for 525^2).
Thus, the valid factors are combinations of 2k⋅522^k \cdot 5^2, where k=0,2,4,6k = 0, 2, 4, 6. These are 4 factors.
Answer: (d) 5