When oil and water are stirred together, water molecules will attract each other electrically, but they will not attract oil molecules, causing separation.
- Water is a polar molecule while oil is a non-polar molecule. Water can only dissolve polar molecules and hence, oil is insoluble in water.
- When water comes in contact with oil, water molecules attract each other, while the oil molecules stick together, hence, forming two separate layers of water and oil. The oil layer floats on the water layer.
- Therefore, when oil and water are stirred together, water molecules will attract each other electrically, but they will not attract oil molecules, causing separation.
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Answer:
Part A: 5.899x10^-3 moles of Al
Part B: 1.573 g of AlBr3
Explanation:
Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.
0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:
Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:
2Al + 3Br2 → 2AlBr3
We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:
of Al
(9) Mechanical advantage = force by machine / force applied to machine
0.6 = 600 / F
F = 1000 N
(2) Efficiency = (output / input) x 100
Efficiency = (500 / 2000) x 100
Efficiency = 25%
(4) The overall energy conversion in power plants is chemical to electrical while in dams it is potential to electrical.
(5) Using the formula:
1568 = 40 x 9.81 x h
h = 4.0 m
(6) Potential to electrical
(10) An object raised and held stationary above the ground.
Answer:
22.5mL is the volume of the water
Explanation:
When the graduated cylinder is in the 25.0mL mark, the mass of this volume is 22.4g. To convert this mass to volume we need to use density, as follows:
22.4g × (1mL / 0.99704g) = 22.5mL is the volume of the water.
That means the cylinder is uncalibrated in 2.5mL when the cylinder is in the 25.0mL mark
Chlorous Acid Ionizes as,
HClO₂ ⇆ H⁺ + ClO₂⁻
Ka = [H⁺] [ClO₂⁻] / [HClO₂]
Ka of Chlorous Acid = 1.1 × 10⁻²
The concentration of H⁺ ions at equilibrium are calculated as,
Initial 0.1 M ⇆ 0 0
At Equilibrium 0.1-X ⇆ X X
So,
Ka = [X] [X] / [0.1-X]
Putting value of Ka,
1.1 × 10⁻² = X² / 0.1-X
Solving for X,
(www.cymath.com)
X = 0.028 M
Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.
Percentage Ionization is calculated as,
= [H⁺] / [HA] × 100
= (0.028 / 0.1) × 100
= 0.28 × 100
= 28 % (Percentage Ionization)
