Answer:
usually the perfumes are made of aromatic hydrocarbons invloving
cetone, ethanol, benzaldehyde, formaldehyde, limonene, methylene chloride, camphor, ethyl acetate, linalool and benzyl alcohol. which have density lower than the water hence they will float on the top of the water.
Hope this helps you
Explanation:
Hi!
The radical bromination reaction of C₆H₅CH₂CH₃ is performed through a mechanism in which radical reactions are involved. This compound is an alkylbenzene compound, and the carbon that is more reactive towards radical bromination is the carbon bonded to the aromatic ring because in the reaction mechanism the intermediaries are stabilized by resonance in the aromatic ring.
A terminal substitution will not occur because substitution there will not be stabilized by resonance. The compound that will be formed in this reaction would be:
C₆H₅CH₂CH₃ + Br₂ → C₆H₅CH₂(Br)CH₃ + HBr
Mass of lead (II) chromate is 51 g. The molecular formula is
and its molar mass is 323.2 g/mol
Number of moles can be calculated using the following formula:

Here, m is mass and M is molar mass.
Putting the values,

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.
Answer:
This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.
Explanation:
Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.
Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.
If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.
Answer: 19.4 mL Ba(OH)2
Explanation:
H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)
At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.
0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2
Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.
0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl
HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.
0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2