because it can influence how frequently and sufficiently the particles collide depending on the space it has to do so, for example a large surface area would be have a slower rate of reaction and a lower temperature. (the rate of reaction in terms of concentration, it is diffused from high to low)
Although the data for the experiment was not provided, we can offer a generalized answer in that when performing an experiment to achieve absolute zero temperatures, the value will never match the exact value.
<h3 /><h3>What is absolute zero?</h3>
Absolute zero is the lower limit of temperature. It is considered the coldest possible temperature that can exist. However, any attempt to reach this temperature in a controlled environment has failed, <u>scientists do not think it is possible to recreate this </u><u>temperature</u><u>. </u>
Therefore, we can confirm that the value of the absolute zero experiments did not match the accepted value. If the hypothesis was that it would be difficult or impossible to achieve, then the data would support the hypothesis, otherwise, it would fail to do so.
In summary, absolute zero is a temperature that cannot be recreated in a lab, so the value in this experiment does not match the accepted value and there is <u>no further exploration </u>to be done on this matter.
To learn more about absolute zero visit:
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It’s the 3rd one obviously bro
Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is
i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
Answer:
There were 0.00735 moles Pb^2+ in the solution
Explanation:
Step 1: Data given
Volume of the KI solution = 73.5 mL = 0.0735 L
Molarity of the KI solution = 0.200 M
Step 2: The balanced equation
2KI + Pb2+ → PbI2 + 2K+
Step 3: Calculate moles KI
moles = Molarity * volume
moles KI = 0.200M * 0.0735L = 0.0147 moles KI
Ste p 4: Calculate moles Pb^2+
For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+
For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+
There were 0.00735 moles Pb^2+ in the solution
