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For Chemistry, plan how both of you will see the equation, graph or structure being discussed. Microsoft Teams is Latimer’s default lesson platform; you and the tutor can agree another platform. Test the arrangement before the first paid lesson. If the learner is under 18, agree communication boundaries and a suitable study space with a parent or carer, with an adult within earshot.

Here is a constructed practice question: 25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid exactly neutralises 20.0 cm³ of sodium hydroxide solution. Find the sodium hydroxide concentration.

The balanced equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The coefficient 2 means that each mole of sulfuric acid reacts with two moles of sodium hydroxide.

1. Convert the acid volume: 25.0 cm³ ÷ 1,000 = 0.0250 dm³.
2. Use amount = concentration × volume: 0.100 × 0.0250 = 0.00250 mol of H₂SO₄.
3. Apply the equation ratio: 2 × 0.00250 = 0.00500 mol of NaOH.
4. Convert the alkali volume to 0.0200 dm³, then divide: 0.00500 ÷ 0.0200 = 0.250 mol dm⁻³.

An answer of 0.125 mol dm⁻³ can result from missing the 2:1 ratio. Seeing each line online lets the discussion focus on that chemical decision, rather than simply correcting the final number.

For a follow-up, keep the acid amount unchanged but suppose it neutralises 40.0 cm³ of sodium hydroxide. The alkali still contains 0.00500 mol, now in 0.0400 dm³, so its concentration is 0.125 mol dm⁻³. Twice the volume is needed because this alkali is half as concentrated. Explain that relationship before reaching for the calculator.

Consider this statement about a reversible reaction at equilibrium: “Adding a catalyst increases the equilibrium yield because the forward reaction is faster.” The missing step is to consider the reverse reaction.

A catalyst speeds up both directions. It does not change the equilibrium position, so the equilibrium yield is unchanged at the same conditions. If the mixture has not yet reached equilibrium, the catalyst helps it get there sooner. A better response is: “A catalyst increases the rates of both forward and reverse reactions. It does not change the equilibrium composition.”

This distinction matters when a question asks separately about rate and yield. In an online discussion, mark which part of your answer concerns speed and which concerns the final mixture. Then try a fresh question without those prompts. For a course that includes equilibrium constants, extend the explanation to why the catalyst does not change the constant at a fixed temperature.

Use Latimer’s free introductory meeting to discuss goals, teaching approach and arrangements. It is normally a conversation about fit, rather than a full teaching lesson. You could bring the question you most wanted help with and explain where you became stuck.