Solid equilibria and adsorption
Session 4: Independent study during the ECS meeting · about 3 hours
Your independent lesson
Your action: Complete the video opening and R1–R3 first, or use the written alternative. Then open the worksheet and Lab guide. Keep attempts and corrections.
This lesson replaces our live class while the instructor attends the ECS meeting.
- Practice worksheet, 6 pages: A–G and I required; H1–H4 optional.
- Lab guide, 4 pages: settings, controls and interpretation.
- Worked solutions: open after your own attempt.
Watch the Lab 10 and Lab 11 demonstrations, then repeat the investigations independently. Save your calculations and study JSON files.
Study route · Complete video slides · Core slide reference · Optional background
The study route
Your action: Keep R1–R3 with your worksheet. Record your actual study time. Do not repeat the written opening after watching it.
- Opening: watch once and write R1–R3 during its four-minute pause.
- A–D: processes, solubility, inventory and Lab 10.
- Break: take 10 minutes away from the screen.
- E–G: adsorption, vessel balance and Lab 11.
- I: submit the exit ticket and feedback once.
Core work totals 100 minutes. Together with the four-minute R1–R3 pause, required work totals 104 minutes, plus the 10-minute break and combined recording. The video page reports measured recording and planned total times, approximately three hours overall.
H1–H4 is optional and adds about 20 minutes if you choose it. These allocations are study activities, not video timestamps.
Process predictions
Your action: Pause for about 8 minutes. Complete A1–A3, worksheet p. 1, before viewing the next slide.
A1. Classify dissolution of a crystal, sublimation into a vessel, and adsorption on a porous solid. Identify whether the process transfers material between bulk phases or changes surface uptake.
A2. Week 3 used Gibbs energy at fixed T,P. Which thermodynamic potential is minimized for a closed, rigid vessel at fixed T,V?
A3. In an ideal gas mixture, distinguish the partial pressure of the subliming species from total pressure. Predict how an inert gas contributes.
For each answer, name the process, the constraints, and one quantity that a material balance must determine.
Process check
Your action: Check A1–A3. Keep your first answers and record the reason for any change.
| Process | Equilibrium description | Amount information |
|---|---|---|
| Crystal and liquid | Solid–liquid equilibrium, SLE | Dissolved and remaining solid |
| Solid and vapor | Gas–solid equilibrium, GSE | Gas and remaining solid |
| Surface uptake | Adsorption equilibrium | Gas and adsorbed material |
At fixed T,P, equilibrium minimizes Gibbs energy. At fixed T,V, it minimizes Helmholtz energy, subject to conserved inventories.
For ideal gases, P_{tot}=p_A+p_{inert}. Lab 10 treats nonreacting bulk equilibria. Lab 11 treats adsorption. A gas–solid chemical reaction would require species and reaction constraints beyond these demonstrations.
Pure-solid standard state
Your action: Read the two phase descriptions. Use them to set up B1, worksheet p. 1.
For component i in a liquid and its present pure solid:
\mu_i^L=\mu_i^S,\qquad \mu_i^L=\mu_i^{L,*}+RT\ln(x_i\gamma_i)
The liquid reference \mu_i^{L,*} is the subcooled pure liquid at the same temperature and pressure.
The present pure solid has unit activity on its pure-solid standard state. Unit activity does not mean zero chemical potential or a zero amount of solid.
x_i\gamma_i specifies a liquid activity. A material balance is still needed to determine phase amounts.
Solubility equation
Your action: Identify the units and assumptions before substituting into B1.
For a present pure solid, with constant fusion enthalpy and negligible fusion heat-capacity correction:
\ln(x_i\gamma_i)=-\frac{\Delta H_{fus,i}}{R} \left(\frac{1}{T}-\frac{1}{T_{m,i}}\right)
Use kelvin, \Delta H_{fus} in J mol^{-1}, and R=8.314462618 J mol^{-1} K^{-1}.
For T<T_m and positive \Delta H_{fus}, the right side is negative. With an ideal liquid, \gamma_i=1 and the result gives x_i^{sat}<1.
This model excludes solid solutions, chemical reactions and heat-capacity corrections. Other solid forms or transitions require additional data.
Solubility attempt
Your action: Pause for about 6 minutes. Complete B1–B2, worksheet p. 1. Calculate before advancing.
Synthetic inputs:
T_m=350\ \mathrm{K},\qquad \Delta H_{fus}=10000\ \mathrm{J\,mol^{-1}},\qquad T=300\ \mathrm{K}
B1. Assume an ideal liquid. Calculate the saturated liquid mole fraction x^{sat} using the equation on slide 7.
B2. Does this number give the moles of solid remaining? State what additional information you would need.
Before calculating, predict whether the saturated mole fraction is above or below 1.
Solubility check
Your action: Check B1–B2. Correct units and distinguish composition from phase amount.
x^{sat}=\exp\left[-\frac{10000}{8.314462618} \left(\frac{1}{300}-\frac{1}{350}\right)\right] =0.563986
The result is a liquid mole fraction, not moles of dissolved solute or remaining solid.
With a nonideal liquid, the equation constrains x\gamma(x,T). The composition must satisfy that relation and the admitted phase model.
Phase amounts require total composition and inventory balances. A binary eutectic adds a second solid branch; it is optional background for this lesson.
Sublimation equilibrium
Your action: Read the condition on solid presence. Begin C1, worksheet p. 2 by identifying the gas species.
For a present pure bulk solid A in equilibrium with its vapor:
\mu_A^S=\mu_A^g
With an ideal gas, negligible solid pressure correction, and the stated sublimation-pressure model:
p_A=P_{sub}(T)\quad\text{while solid remains}
p_A is the partial pressure of the subliming species. An inert gas contributes to total pressure.
If all solid disappears, there is no remaining bulk-solid phase to enforce this equality. The finite inventory may give p_A<P_{sub}.
Finite-vessel inventory
Your action: Use this decision sequence for C1–C3. Write the capacity before deciding which phases are present.
For a closed, rigid vessel at fixed T,V_g with ideal gases:
n_g^{sat}=\frac{P_{sub}V_g}{RT},\quad n_A^g=\min(n_A^{tot},n_g^{sat}),\quad n_A^s=n_A^{tot}-n_A^g
p_A=\frac{n_A^gRT}{V_g},\qquad P_{tot}=p_A+\frac{n_{inert}RT}{V_g}
Use R=8.314462618 kPa L mol^{-1} K^{-1} with pressure in kPa and gas volume in L.
For C3, (\mu_A^g-\mu_A^S)/(RT)=\ln(p_A/P_{sub}). A present solid requires zero. An exhausted-solid state may have a negative value.
Inventory worked example
Your action: Follow the arithmetic, then cover these results before attempting worksheet C.
Synthetic case: T=300 K, V_g=1 L, P_{sub}=10 kPa.
n_g^{sat}=\frac{10(1)}{8.314462618(300)}=0.00400908\ \mathrm{mol}
| n_A^{tot} (mol) | n_A^g (mol) | n_A^s (mol) | p_A (kPa) |
|---|---|---|---|
| 0.010 | 0.00400908 | 0.00599092 | 10.0000 |
| 0.002 | 0.00200000 | 0 | 4.98868 |
Adding 0.002 mol inert gas at the same T,V_g adds 4.98868 kPa to total pressure in either case. It leaves p_A and the inventory threshold unchanged in this model.
Inventory attempt
Your action: Pause for about 10 minutes. Complete C1–C3, worksheet p. 2, before viewing slide 14.
Use a new synthetic vessel: T=300 K, V_g=2 L and P_{sub}=8 kPa.
C1. Calculate n_g^{sat}. Find gas moles, solid moles and p_A for n_A^{tot}=0.010 mol and 0.004 mol.
C2. Add n_{inert}=0.001 mol to each case at the same T,V_g. Calculate P_{tot} and identify which results change.
C3. Calculate \ln(p_A/P_{sub}) for both cases. Explain why zero is required when solid remains, while a negative value can be a valid exhausted-solid boundary state.
Use R=8.314462618 kPa L mol^{-1} K^{-1}. Show substitutions and units.
Inventory check
Your action: Check C1–C3. Keep enough digits in the capacity before subtracting the remaining solid.
n_g^{sat}=0.00641453\ \mathrm{mol},\qquad p_{inert}=1.24717\ \mathrm{kPa}
| n_A^{tot} (mol) | n_A^g (mol) | n_A^s (mol) | p_A (kPa) | P_{tot} (kPa) |
|---|---|---|---|---|
| 0.010 | 0.00641453 | 0.00358547 | 8.00000 | 9.24717 |
| 0.004 | 0.00400000 | 0 | 4.98868 | 6.23585 |
The first case retains solid and has \ln(p_A/P_{sub})=0. The second exhausts it and has \ln(p_A/P_{sub})\approx-0.472271.
In each row, n_A^g+n_A^s=n_A^{tot}. The inert changes total pressure, while p_A is unchanged at fixed T,V_g in this model.
Lab 10 predictions
Your action: Pause for about 5 minutes. Complete D1, worksheet p. 3. Read Lab guide pp. 1–2, but predict before running the Lab.
Select GSE · sublimation in a rigid vessel at T=300 K and V_g=1 L. Set the sublimation reference pressure to 10 kPa at 300 K.
| Case | Subliming-species inventory (mol) | Inert gas (mol) |
|---|---|---|
| 10A | 0.010 | 0 |
| 10B | 0.002 | 0 |
| 10C | 0.002 | 0.002 |
Predict solid presence, p_A and total pressure. Identify the single input changed between successive cases.
Lab 10 demonstration
Your action: Watch the actual Lab 10 demonstration. Follow the controls using Lab guide pp. 1–2. Keep D1 predictions visible.
The demonstration follows this route:
- Select the gas–solid vessel mode and enter the reference state.
- Compare inventories above and below the solid-depletion threshold.
- Add inert gas and compare partial pressure with total pressure.
- Read the plots and check the species inventory balance.
- Save study JSON, reopen it, verify inputs, and calculate again.
Record results during your independent rerun at slide 18. Verify restored controls and calculate again before trusting a displayed result.
Lab 10 evidence
Your action: Use Lab guide pp. 1–2 to identify the evidence you will retain for D2–D4.
| Read or save | Explain what it establishes |
|---|---|
| Inputs and active mode | Temperature, volume and inventory constraints |
| Solid remaining and gas amount | Which phases are present at the selected point |
| Partial and total pressure | Subliming-species and inert contributions |
| Inventory check | Gas plus solid equals the subliming-species total |
| Study JSON | A reproducible parameter and result record |
The actual volatile partial pressure follows P_{sub}(T) wherever solid remains. The plots sweep temperature, not time. A plotted boundary does not mean every possible phase is present at the selected state.
Lab 10 independent investigation
Your action: Pause for about 22 minutes. Complete D2–D4, worksheet p. 3, using Lab guide pp. 1–2 and Lab 10.
D2. Run cases 10A, 10B and 10C from D1. Record phase presence, gas and solid moles, partial pressure and total pressure.
D3. Check gas capacity, the species inventory and the inert contribution by hand. Use the plots to explain the transition from solid present to solid exhausted.
D4. Export the cases. Reopen a JSON file, restore the controls, verify them, and recalculate. Record whether the results reproduce.
Retain one comparison table, your calculations and the saved files. Explain any discrepancy before advancing.
Lab 10 check
Your action: Compare D2–D4 with these values and the worked solutions. Correct the reasoning as well as the arithmetic.
At 300 K and 1 L, the gas capacity is 0.00400908 mol.
| Case | p_A (kPa) | Solid (mol) | P_{tot} (kPa) |
|---|---|---|---|
| 10A: 0.010 mol A, no inert | 10.0000 | 0.00599092 | 10.0000 |
| 10B: 0.002 mol A, no inert | 4.98868 | 0 | 4.98868 |
| 10C: 0.002 mol A, 0.002 mol inert | 4.98868 | 0 | 9.97736 |
A pressure below P_{sub} is the expected finite-inventory result after solid depletion. It is not a failed saturation calculation.
These are equilibrium endpoints. They do not give a sublimation rate or the time needed to reach equilibrium.
Break
Your action: Pause for 10 minutes. Save your Lab 10 files and leave the screen briefly.
When you return, open worksheet p. 4 and Lab guide pp. 3–4.
The next part treats gas uptake on a surface. You will calculate an isotherm value, then solve an inventory balance for a closed vessel.
The same question remains useful: What does equilibrium determine, and what must the material balance determine?
Adsorption and bulk phases
Your action: Read the definition of uptake before starting E, worksheet p. 4.
Adsorption describes material associated with a surface. An isotherm gives uptake at a specified temperature and gas pressure.
q(P,T)=\frac{\text{moles adsorbed}}{\text{mass of adsorbent}}
Here, q is absolute uptake in mol kg^{-1} and P is the adsorbate partial pressure in kPa.
Adsorbed moles are n_{ads}=m_s q, with adsorbent mass m_s in kg. The symbol q itself is not an amount in moles.
Bulk sublimation and surface adsorption require different state descriptions and inventory balances.
Langmuir model
Your action: Identify the parameters and dimensionless group. Use this expression for E1–E3.
q=q_s\frac{bP}{1+bP},\qquad \theta=\frac{q}{q_s}=\frac{bP}{1+bP}
| Quantity | Meaning | Unit used here |
|---|---|---|
| q_s | Saturation uptake | mol kg^{-1} |
| b | Affinity parameter | kPa^{-1} |
| bP | Dimensionless pressure group | 1 |
| \theta | Fractional coverage in the model | 1 |
The idealized model assumes a finite set of equivalent sites and no lateral interactions. A good fit alone does not establish this microscopic mechanism.
Henry limit
Your action: Derive the low-pressure slope from Langmuir. Retain the units when comparing the models.
If bP\ll1:
q\approx q_s bP=K_H P,\qquad K_H=q_s b
K_H has units mol kg^{-1} kPa^{-1} for this adsorption convention.
The matched Henry line and Langmuir curve share the slope at P\rightarrow0. The Henry line has no saturation plateau and requires a low-pressure domain.
A gas-solubility convention such as p=Hx defines a different constant with different units. State the equation and concentration basis whenever using the name “Henry constant.”
Isotherm worked example
Your action: Follow this calculation, then use the same parameters for E1–E3.
Synthetic parameters at fixed T: q_s=2 mol kg^{-1} and b=0.1 kPa^{-1}.
q=\frac{2(0.1)P}{1+0.1P},\qquad K_H=0.2\ \mathrm{mol\,kg^{-1}\,kPa^{-1}}
| P (kPa) | bP | Langmuir q (mol kg^{-1}) | Henry q (mol kg^{-1}) |
|---|---|---|---|
| 1 | 0.1 | 0.181818 | 0.200000 |
| 10 | 1 | 1.000000 | 2.000000 |
At P=1/b=10 kPa, Langmuir coverage is one half. At this pressure, a low-pressure linear approximation is already poor.
Isotherm attempt
Your action: Pause for about 10 minutes. Complete E1–E3, worksheet p. 4. Calculate before viewing slide 26.
Keep q_s=2 mol kg^{-1} and b=0.1 kPa^{-1}. Use matched Henry K_H=q_s b=0.2 mol kg^{-1} kPa^{-1} for comparison.
E1. Calculate Langmuir and matched Henry uptake at 5 kPa and 40 kPa. Also calculate Langmuir fractional coverage.
E2. Calculate bP at each pressure and compare with the low-pressure condition bP\ll1.
E3. Predict each model’s high-pressure behavior. Which model can exceed q_s mathematically, and what does that mean for its use?
Show the substitutions and units. Compare your calculations with the matched Henry values on the next slide.
Isotherm check
Your action: Check E1–E3. Explain model disagreement using bP, not only curve shapes.
| P (kPa) | bP | Langmuir q | \theta | Henry q |
|---|---|---|---|---|
| 5 | 0.5 | 0.666667 | 0.333333 | 1.00000 |
| 40 | 4 | 1.60000 | 0.800000 | 8.00000 |
Uptake values are in mol kg^{-1}. As pressure increases, Langmuir uptake approaches 2 mol kg^{-1}. The Henry line continues increasing without a plateau.
At 40 kPa, bP=4 is far outside bP\ll1. The large Henry value shows why a low-pressure approximation needs a stated domain.
Closed adsorption vessel
Your action: Read the balance before F1–F3, worksheet p. 4. Identify which pressure is unknown.
For an ideal gas and absolute uptake in a closed vessel:
n_A^{tot}=\underbrace{\frac{PV_g}{RT}}_{n_A^g} +\underbrace{m_s q(P,T)}_{n_A^{ads}}
Use the accessible gas volume V_g and adsorbent mass m_s in kg. The unknown P is adsorbate partial pressure.
At fixed T,V_g,m_s,n_A^{tot}, the isotherm and balance determine P together. Both terms increase with P\ge0 for positive Langmuir parameters.
Excess uptake requires a gas-density and volume convention. It cannot be substituted directly for absolute uptake in this balance.
Vessel worked example
Your action: Follow this known-pressure calculation. Use its inventory to justify your predictions in F1–F3.
Synthetic case at 300 K, 1 L and 10 kPa: use m_s=10 g, q_s=2 mol kg^{-1} and b=0.1 kPa^{-1}.
q=2\frac{0.1(10)}{1+0.1(10)}=1\ \mathrm{mol\,kg^{-1}}
n_A^g=\frac{10(1)}{8.314462618(300)}=0.00400908\ \mathrm{mol}
n_A^{ads}=0.010(1)=0.0100000\ \mathrm{mol},\quad n_A^{tot}=0.01400908\ \mathrm{mol}
The next task uses a smaller total inventory, 0.010 mol, with the same other parameters. That change will require a different equilibrium pressure.
Lab 11 predictions
Your action: Pause for about 5 minutes. Complete F1–F3, worksheet p. 4. Read Lab guide pp. 3–4 before watching.
Use T=300 K, V_g=1 L, n_A^{tot}=0.010 mol, q_s=2 mol kg^{-1}, b=0.1 kPa^{-1} and initially 10 g adsorbent.
F1. Compared with slide 28, predict whether Langmuir pressure is above or below 10 kPa. Explain using the smaller inventory.
F2. Switch to matched Henry adsorption, K_H=0.2 mol kg^{-1} kPa^{-1}. Predict which model gives the lower pressure at this fixed inventory.
F3. Return to Langmuir and double adsorbent mass to 20 g. Predict the pressure change.
Lab 11 demonstration
Your action: Watch the actual Lab 11 demonstration with Lab guide pp. 3–4. Compare results with your F1–F3 predictions.
The demonstration follows this route:
- Set pressure and uptake units, model parameters and temperature.
- Compare the Langmuir curve with its matched Henry limit.
- Enter the closed-vessel inventory and accessible gas volume.
- Compare 10 g and 20 g of adsorbent and check the inventory balance.
- Export a study, reopen it, verify the controls, and calculate again.
Use the independent rerun at slide 32 to build your own comparison. Record the selected model and parameter units with every result.
Adsorption evidence
Your action: Prepare the columns for G1–G4, worksheet p. 5, using Lab guide pp. 3–4.
| Record | Why it matters |
|---|---|
| Model, units and temperature | Parameters depend on the equation and convention |
| Pressure and uptake | An isotherm point is a state, not a total amount |
| Gas moles and adsorbed moles | Their sum must reproduce the closed inventory |
| Adsorbent mass in kg | m_s q converts uptake to moles |
| Saved settings and recalculated output | The comparison can be reproduced |
Shared low-pressure slope does not make models identical over a finite range. A good inventory balance also does not establish a model’s physical validity.
Lab 11 independent investigation
Your action: Pause for about 24 minutes. Complete G1–G4, worksheet p. 5, using Lab guide pp. 3–4 and Lab 11.
G1. Run 11A: Langmuir, 10 g; 11B: Henry, 10 g; and 11C: Langmuir, 20 g. Keep T=300 K, V_g=1 L and n_A^{tot}=0.010 mol. Use q_s=2, b=0.1 and matched K_H=0.2 in the stated units.
G2. Record pressure, uptake, gas moles and adsorbed moles. Check each inventory balance independently.
G3. Compare with your predictions and inspect the graph. Explain model choice and mass effects, including the low-pressure limitation.
G4. Save the studies, reopen one, restore and verify settings, then recalculate. Record the reproduced result.
Lab 11 check
Your action: Check G1–G4 against these synthetic results. Use unrounded values for the balance.
| Case | P (kPa) | q (mol kg^{-1}) | n_A^g (mol) | n_A^{ads} (mol) |
|---|---|---|---|---|
| 11A: Langmuir, 10 g | 6.08025 | 0.756238 | 0.00243762 | 0.00756238 |
| 11B: Henry, 10 g | 4.16509 | 0.833018 | 0.00166982 | 0.00833018 |
| 11C: Langmuir, 20 g | 2.84498 | 0.442971 | 0.00114057 | 0.00885943 |
Each row sums to 0.010 mol. More adsorbent lowers pressure and increases adsorbed moles, even though uptake per kilogram decreases.
The Henry result illustrates model sensitivity. The finite value of bP requires caution about treating it as a valid low-pressure approximation.
Model scope
Your action: Read these limitations. Choose one that matters to your interpretation. Extensions are optional.
| Extension | Required qualification |
|---|---|
| Freundlich | Empirical pressure range; no saturation plateau |
| BET | Relative pressure P/P_0, a fitting range and model assumptions |
| Isosteric heat Q_{st} | Pressures at the same absolute loading across temperatures |
| Experimental fitting | Data, uncertainty, units and a defined fitting objective |
Today’s tasks use synthetic parameters. They do not establish a material’s adsorption mechanism or validate extrapolation.
The required work is the model comparisons and inventory checks. No new fit, BET calculation or isosteric-heat calculation is required.
Optional unseen-case practice
Your action: For the required route, go to exit ticket and feedback. Optional H1–H4 on worksheet p. 6 adds practice; attempt it before opening the solutions. Use slides 11, 22 and 27.
GSE: T=310 K, V_g=2 L and P_{sub}=5 kPa. Compare total inventories of 0.002 mol and 0.008 mol of subliming species. Add 0.001 mol inert gas to each.
Adsorption: A different synthetic Langmuir material has q_s=3 mol kg^{-1} and b=0.05 kPa^{-1}. At T=310 K and P=20 kPa, use V_g=2 L and 5 g adsorbent.
Calculate the state and amount quantities. Then predict the pressure change if adsorbent mass doubles at the same total inventory, temperature and accessible gas volume.
Choose whether to attempt H
Your action: Choose your route. Continue to exit ticket and feedback to finish the required lesson. If you choose optional H1–H4, worksheet p. 6, allow about 20 extra minutes and attempt it before slide 37. No mandatory H pause is required.
H1. Calculate the GSE gas capacity. Decide whether solid remains for each inventory and find gas moles, solid moles and p_A.
H2. Include the inert gas. Calculate total pressure and check each species inventory.
H3. For adsorption, calculate uptake, gas moles, adsorbed moles and the required total inventory at 20 kPa.
H4. Predict the effect of doubling adsorbent mass at fixed inventory. Explain using the balance, then state one shared assumption and one difference between bulk GSE and adsorption. An exact new pressure is not required.
Unseen check
Your action: Check H1–H4. Keep your original attempt and annotate corrections. The solutions show intermediate calculations.
GSE gas capacity: 0.00387975 mol. Inert contribution: 1.28874 kPa.
| n_A^{tot} (mol) | n_A^g (mol) | n_A^s (mol) | p_A (kPa) | P_{tot} (kPa) |
|---|---|---|---|---|
| 0.002 | 0.00200000 | 0 | 2.57748 | 3.86623 |
| 0.008 | 0.00387975 | 0.00412025 | 5.00000 | 6.28874 |
Adsorption: q=1.5 mol kg^{-1}, gas 0.0155190 mol, adsorbed 0.0075000 mol, total 0.0230190 mol.
Doubling adsorbent mass at fixed inventory lowers pressure: at the old pressure, the balance would contain too many moles.
Choosing an equilibrium description
Your action: Use these questions to prepare your reflection in I, worksheet p. 6.
- What process occurs? Bulk phase transfer, surface uptake or chemical reaction?
- What is fixed? Temperature, pressure, accessible volume and/or species inventory?
- Which model and convention apply? Standard states, gas ideality, uptake basis, units and pressure range?
- What sets amounts? Combine the equilibrium relation with material balances and check phase presence.
A solubility composition, sublimation pressure or isotherm loading describes part of the equilibrium state. None replaces an inventory balance.
Report a numerical result with its assumptions and a check that supports the interpretation.
Exit ticket and feedback
Your action: Pause for about 10 minutes. Complete I, worksheet p. 6, then submit the Session 4 form once.
Explain what happens to gas pressure when bulk solid runs out. In the evidence answer, include one Lab 10 comparison and one Lab 11 numerical balance, with settings and units.
Report your actual study time, access problems, and what would improve the next lesson.
The form asks for no name, student ID or email. Wait for “Response received.”
Receipt and saved work
Your action: Verify the submission receipt and save your work. If the form shows “Response received,” do not submit again.
Keep these records together:
- Completed worksheet, including predictions, calculations and corrections.
- Lab 10 and Lab 11 comparison tables and study JSON exports.
- One reproduced result after reopening and recalculating a saved study.
- Your actual study time and confirmation that you submitted the form.
The recording, worksheet and Lab investigations form this week’s lesson. There is no additional compulsory after-class task.