Answer:
d. Adding 5% HCl solution to the crude reaction mixture will protonate aniline increasing its solubility in the aqueous solution.
Explanation:
On this case, we have to check the <u>structures of each compound</u> (figure 1). For naphthalene we dont have <u>any functional groups</u> therefore, the addition of HCl or NaOH it will not affect naphthalene so <u>we can discard "B" and"C".</u>
When we add HCl solution we will have the production 
the presence of this <u>hydronium ion will protonate the acid</u>, so we can <u>discard a.</u>
<u />
Finally, for d when we add the <u>hydronium ion will react with aniline</u> (a base) and will produce an <u>ammonium ion</u>. This ammonium ion have a <u>positive charge</u>, therefore the <u>polarity will increase</u> and the molecule would be more soluble on water (figure 2).
I hope it helps!
Answer:
CHARLES' LAW
given:
= 600 mL = 0.6 L
= 27 °C = 300.15 L
= 77 °C = 350.15 L
conversion:
= 600 mL (1 L / 1000 mL)
= 0.6 L
= 27 °C + 273.15 K
= 300.15 K
= 77 °C + 273.15 K
= 350.15 K
solution:
= ( × ) ÷
= (0.6 L × 350.15 K) ÷ 300.15 K
= 0.7 L
Think it though and always go back to the cause of the problem
Soil is built up of clastic particles, organics, and clay while sand is a combination of rocks, minerals, and fragments of rotting shells.
Soil particles are larger in size and mass together, Sand particles are tinier and individual.
<u>Explanation:</u>
So, the soil is weaker than sand particles. When we mix soil and sand into the water, we acknowledge that soil particles somewhat settle at the bottom while designing parts mixed into water. Sand is almost heavier than water so it settles down at the base.
Soil particles settle faster and will be on the bottom while the sand particles will be on top. And no this is not a great method to define particle size distribution.
1. Nucleic
2. Protein
3. Carbohydrates
4. Lipids
