Class prep · TJ Honors Chemistry

Understand chemistry,from the first atom to electrochemistry.

A full honors chemistry course taught with worked examples and original practice, aligned in depth to the front half of AP Chemistry and pitched for TJ-level reasoning. Every problem has a full worked solution. Built by Neil Kothari, a TJ alumnus.

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$395 covers the full school year. The first lesson is free.

Try one. This is a test-level problem from the course.

Problem 1

Test level
Multiple choice
No calculator
4 pts

A rectangular tank measures $2.0\,\mathrm{m}$ by $50\,\mathrm{cm}$ by $30\,\mathrm{cm}$ and is completely filled with a liquid of density $0.90\,\mathrm{g/cm^3}$ . Which value is closest to the mass of the liquid? No calculator is needed.

Answer

$2.7\times10^{2}\,\mathrm{kg}$ , because the volume is $200\times50\times30=3.0\times10^{5}\,\mathrm{cm^3}$ , the mass is $0.90\times3.0\times10^{5}=2.7\times10^{5}\,\mathrm{g}$ , and dividing by $1000$ gives kilograms.

Full solution

Put every length in the same unit before multiplying, and choose the unit that matches the density. The density uses $\mathrm{cm^3}$ , so use centimetres: $2.0\,\mathrm{m}=200\,\mathrm{cm}$ .
Volume: $200\times50\times30=200\times1500=3.0\times10^{5}\,\mathrm{cm^3}$ .
Mass in grams: density is grams per $\mathrm{cm^3}$ , so multiply: $0.90\,\dfrac{\mathrm{g}}{\mathrm{cm^3}}\times3.0\times10^{5}\,\mathrm{cm^3}=2.7\times10^{5}\,\mathrm{g}$ .
Convert to kilograms: $2.7\times10^{5}\,\mathrm{g}\times\dfrac{1\,\mathrm{kg}}{10^{3}\,\mathrm{g}}=2.7\times10^{2}\,\mathrm{kg}$ , choice (E).
Reasonableness: the tank holds about $0.3\,\mathrm{m^3}$ , which is about $300$ litres, and a liquid slightly lighter than water has a mass slightly under $300\,\mathrm{kg}$ . That agrees with $270\,\mathrm{kg}$ .
The wrong choices are each a clean slip. Choice (A) multiplies $2.0\times50\times30$ with the metres not converted, giving $3000\,\mathrm{cm^3}$ and $2.7\,\mathrm{kg}$ . Choice (B) divides the volume by the density instead of multiplying. Choice (C) stops at grams and calls them kilograms. Choice (D) multiplies by $10^{3}$ instead of dividing when converting grams to kilograms.

Where students lose points

Mixing units inside one product. Convert every length to the unit used in the density before multiplying, and then check that the answer's size matches a quick estimate in the units you want.

original problems, each with a full worked solution: 390; original problems, each with a full worked solution
at test level or beyond: 270; at test level or beyond
short lessons, one for every idea: 30; short lessons, one for every idea
units, all available now: 5; units, all available now

Free lesson

Lesson 1.1: Measuring, Converting and Trusting Numbers

How to read an instrument, carry the right number of digits through a calculation, convert units without slips, and judge how much a result can be trusted.

This lesson starts from zero. No chemistry background is needed.

Problem 2

Core
True or false
No calculator
2 pts

True or False: A balance reads to the nearest $0.01\,\mathrm{g}$ and shows $12.40\,\mathrm{g}$ for a rock. Writing the mass as $12400\,\mathrm{mg}$ correctly shows that the measurement has four significant figures. Justify your answer.

Answer

False. Written as $12400\,\mathrm{mg}$ with no decimal point, the trailing zeros are ambiguous (the number could be read as having three, four or five significant figures). The reading is certain only to about $\pm10\,\mathrm{mg}$ , so only scientific notation, $1.240\times10^{4}\,\mathrm{mg}$ , shows exactly four.

Full solution

Start with the measurement itself. A balance that reads to $0.01\,\mathrm{g}$ gave $12.40\,\mathrm{g}$ : the digits $1$ , $2$ , $4$ are certain and the final $0$ is the estimated digit. All four digits count, including that final zero, because it records a real reading.
Convert with $1\,\mathrm{g}=1000\,\mathrm{mg}$ : $12.40\,\mathrm{g}\times\dfrac{1000\,\mathrm{mg}}{1\,\mathrm{g}}=12400\,\mathrm{mg}$ . The number is right, but the way it is written is the problem.
The uncertainty of $\pm0.01\,\mathrm{g}$ is $\pm10\,\mathrm{mg}$ . So in $12400\,\mathrm{mg}$ the tens digit is the last meaningful one, and the ones digit is only a placeholder. A reader who sees $12400$ with no decimal point cannot tell which zeros are placeholders and which are measured.
The convention is that trailing zeros in a number without a decimal point are not reliably significant. So $12400$ could mean three, four or five significant figures, and the statement fails.
Fix it with scientific notation, where every digit in the coefficient is significant: $1.240\times10^{4}\,\mathrm{mg}$ has exactly four. Changing units should never change the number of significant figures, and scientific notation is how you make that visible.
Check: the coefficient $1.240$ has four digits, the same count as $12.40$ . The conversion only moved the decimal point.

Where students lose points

Assuming a unit conversion by a power of ten keeps the digit count automatically. The count survives only if you write the answer so the trailing zeros are unambiguous. Whenever a converted answer ends in zeros without a decimal point, switch to scientific notation.

Problem 3

Core
Multiple choice
No calculator
4 pts

A student measures the density of a syrup and gets $1.30\,\mathrm{g/mL}$ . The accepted value is $1.25\,\mathrm{g/mL}$ . What is the percent error of the student's value?

Open the rest of this lesson, free

The full lesson and all 13 of its problems, with every solution.

How it works

Class prep that works alongside class

A full honors chemistry course for grades 9 to 12. It works well alongside a school chemistry class and also stands on its own. Students who already know the material get faster and more careful. Students who feel behind rebuild a topic with the lesson before they try the problems.

Every problem is original and pitched for TJ-level reasoning: true or false, multiple choice, multi-part and free response.
Each one is marked core, test level or beyond, and calculator or no calculator.
Solutions show every step and end with the mistake that most often costs points.
Worked examples in every lesson, and diagrams wherever a picture makes the idea clear.

What is inside

1Matter, Measurement and the Atom6 lessons · 78 problems
2Bonding, Structure and Naming6 lessons · 78 problems
3The Mole, Reactions and Stoichiometry6 lessons · 78 problems
4Gases, Energy and Solutions6 lessons · 78 problems
5Rates, Equilibrium, Acids and Bases, Redox6 lessons · 78 problems

The front entrance of Thomas Jefferson High School for Science and Technology

Who built it

Neil Kothari

TJHSST alumnus · Co-founder, EduAvenues · 10+ years in TJ preparation

Neil is a TJ alumnus and has spent more than a decade working with TJ students and their families. He built this course so students can practice the particle-level reasoning and the calculations together. Every problem is written for the course and every answer is checked before it goes in.

Tuition

$395for the full school year

One tuition covers every unit of the course. All 5 units are available now. The Measuring, Converting and Trusting Numbers lesson stays free for everyone.

All 5 units: 30 lessons and 390 original problems
A full worked solution for every problem
Access through the end of the school year

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