Answer:
the second one
Explanation:
im positive bc ur gaining 2 ions
Answer:
393 cm
Explanation:
Step 1: Given and required data
- Density of liquid methylene bromide (ρ): 2.50 g/mL (2.50 g/cm³)
- Earth's gravity (g): 9.81 m/s²
- Atmospheric pressure (P): 0.950 atm
Step 2: Convert 0.950 atm to Pa (N/m²)
We will use the conversion factor 1 atm = 101325 Pa.
0.950 atm × 101325 Pa/1 atm = 9.63 × 10⁴ Pa
Step 3: Convert 2.50 g/cm³ to kg/m³
We will use the conversion factors:

Step 4: Calculate the height (h) of the liquid column
We will use the following expression.
P = ρ × g × h
h = P / ρ × g
h = 9.63 × 10⁴ Pa / (2.50 × 10³ kg/m³) × 9.81 m/s²
h = 3.93 m = 393 cm
Answer:
Explanation:
The objective of this experiment is to match the expected result with each of the propositions given after the experiment had been carried out.
1. The top layer was the <u>organic </u> layer. This is because all the organic compounds have lesser density than water except chloroform that will be formed when NaOH is added.
2. 9-fluorenone was most soluble in the <u>organic </u>layer. This is so as a result of its non-polar carbon structure.
3. Deprotonated chloroanilic acid was most soluble in the <u>aqueous </u>layer as a result of the formation of an electrovalent bond with water.
4. The bottom layer was the <u>aqueous </u>layer as a result of the huge amount of water density.
5. The organic layer was a <u>yellow</u> colored solution.
6. The aqueous layer turned into a <u>pink</u><u> </u> colored solution.
Answer:
Explanation:
Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.
Answer: Mass of silver deposited at the cathode is 37.1g
Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte
Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.
+e− ==> Ag(s)
Given that
Current (I) = 8.5A
Time (t) = 65 *60 = 3900s
Quantity of electricity passed = 8.5*3900 =33150C
Molar mass of Ag= 108g
96500C will liberate 108g
33150C will liberate Xg
Xg= (108*33150)/96500
=37.1g
Therefore the mass of Ag deposited at the cathode is 37.1g.