Answer:
Explanation:
Boyle's law states that:
"For an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume".
Mathematically:
where
p is the pressure of the gas
V is the volume of the gas
The important condition in order to verify Boyle's law in an experiment is that the temperature of the gas must remain constant. This means that the temperature of the gas during the experiment must be kept constant. This can be achieved, for example, by placing the apparatus containing your gas (for example, a glass tube) inside a thermal bath: for instance, a large amount of water kept at constant temperature. This way, the gas inside the tube will also have the same temperature of the water, which is constant.
Learn more about ideal gases:
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Answer:
The plate would only begin to slowly cool down
Explanation:
As the plate is introduced to room temperature the plate kind of stimulates oxidation in a way as in the plate is no longer in a state of continuous heat.
Answer:
Explanation:
Hello there!
In this case, according to the attached solubility chart, it is possible for us to realize that about 88 grams of KNO3 are soluble at 50 °C but just 30 grams are soluble at 20 °C in the same 100 g of water.
In such a way, the crystalized mass of this solute can be calculated by subtracting the mass at 50 °C and the mass at 20 °C:
Best regards!
Aluminum or glass I think
Answer:
<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
Explanation:
- Adding solute to water causes the depression of the freezing point.
<em>ΔTf = Kf.m,</em>
Where,
ΔTf is the change in the freezing point.
Kf is the freezing point depression constant (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>Molality is the no. of moles of solute per kg of the solution.</em>
- <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>
<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>
∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.
<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
