Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula p1V1 = P2V2 , where P1V1 denotes initial pressure and volume and P2V2 denotes values of pressure and volume.
Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p
V = 50.0 p / 2p
= 50.0 /2
= 25.0 m^3
b. if the pressure is cut in half
50.0 p = V x p/2
100 p = V x p
V = 100 m^3
c. if the pressure is tripled
50.0 p = V x 3p
V = 50.0 p / 3p
= 50.0 /3
=16.7 m^3
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The 2 parts or components that make up a solution would be the solute and the solvent.
Answer:
Here's what I get.
Explanation:
Quinine contains phosphors, substances that glow when they are hit with certain wavelengths of light.
The phosphors in quinine absorb UV light, which is invisible to our eyes.
Electrons in the phosphors absorb the UV energy and are excited to higher energy levels.
When the electrons drop back to lower energy levels, they emit some of this energy as a glowing blue visible light.
Answer:
<u><em>Pentane </em></u>
Explanation:
since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it 
Answer:
1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Explanation:
1) Possible reactions
2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Mass of each metal
a) Mass of Cu
The waste was the unreacted copper.
Mass of Cu = 2.5 g
b) Masses of Al and Fe
We have two relations
:
Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g
H₂ from Al + H₂ from Fe = 6.38 L at NTP
i) Calculate the moles of H₂
NTP is 20 °C and 1 atm.

(ii) Solve the relationship
Let x = mass of Al. Then
7.5 - x = mass of Fe
Moles of Al = x/27
Moles of Fe = (7.5 - x)/56
Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x
/18
Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56
∴ x/18 + (7.5 - x)/56 = 0.2652
56x + 18(7.5 - x) = 267.3
56x + 135 - 18x = 267.3
38x = 132.3
x = 3.5 g
Mass of Al = 3.5 g
Mass of Fe = 7.5 g - 3.5 g = 4.0 g
The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g