Answer:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin
Answer:
Carbon 3 is double bonded to an oxygen and attached to carbon 2 and carbon 4. :
Answer: Carbonyl group ( Ketone or aldehyde)
Carbon 17 is attached to an oxygen, which is attached to a hydrogen. :
Answer: Carboxyl group (Carboxylic acid)
A central carbon is attached to an amine, two hydrogens, and a carbon that is double bonded to an oxygen and single bonded to an oxygen attached to a hydrogen. :
Answer: Amide group
An amide group contains both amine and carboxyl
Answer:
The atoms in the first period have electrons in 1 energy level. The atoms in the second period have electrons in 2 energy levels. The atoms in the third period have electrons in 3 energy levels. The atoms in the fourth period have electrons in 4 energy levels.
