Cathode rays are a stream of negatively charged particles or electrons in vacuum tubes. They were first seen by Johann Hittorf in 1869. They are named since they are emitted by electrode or cathode. In order to release the electrons, the electrons should first be detached from the atoms of the cathode.
Answer:
Mass in kg = 4.7*10^19 kg
Mass in tons = 5.2*10^16 tons
Explanation:
<u>Given:</u>
Total volume of sea water = 1.5*10^21 L
Mass % NaCl in seawater = 3.1%
Density of seawater = 1.03 g/ml
<u>To determine:</u>
Total mass of NaCl in kg and in tons
<u>Calculation:</u>
Unit conversion:
1 L = 1000 ml
The volume of seawater in ml is:
To convert mass from g to Kg:
1000 g = 1 kg
To convert mass from g to tons:
1 ton = 9.072*10^6 g
35 m/s = 210000 cm/min
35*600=210000
Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ + H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
O moles - 0.64 mmol
Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol
In menthol
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C H O
0.64 1.28 0.064
x1000 x1000 x1000 to get whole numbers
640 1280 64
10 20 1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O
