What we say:

Magnesium and diatomic oxygen yields magnesium oxide.

Reasoning:

Because we're trying to help learners to learn, it's probably appropriate to be pretty careful about spelling out when we're talking about O2 vs O. And the "and" needs to be there because we need an unambiguous conjunction ("and") that won't be confused with an ion charge ("plus").

What we say:

The reaction of magnesium and diatomic oxygen yields magnesium oxide.

Equation 1: Left side, one magnesium atom, 2 oxygen atoms. Right side, one magnesium atom, one oxygen atom.

Equation 2: Right side multiplied by two results in: Left side, one magnesium atom, two oxygen atoms. Right side, two magnesium atoms, two oxygen atoms.

Etc.

How we read it:

Synthesis reaction: General format, A and B yields A B. Example, 2 M G and O 2 yields 2 M G O

Reasoning:

Not reading formulas here, because we're trying to teach types of reactions. Atomic makeup of compounds is important.

How we read it:

One unit of solid copper and 2 units of aqueous silver nitrate yields 2 units of solid silver and one unit of aqueous silver nitrate.

Reasoning:

We use the generic term "unit" to emphasize that coefficients in equations can be atoms, molecules, ions, formula units, moles, or volumes of gas. We also suggest switching the order, so the states of matter act like adjectives, rather than being read out in the order it is presented in the notation of the equation.

How we read it:

(Top equation)

Forty-five point three moles of diatomic oxygen times 2 moles of magnesium oxide per 1 mole of oxygen, equals ninety-six point 6 moles of magnesium oxide because moles of oxygen cancel.

Reasoning:

Here, we want to emphasize the most important thing first-the result. And then, follow up with the extras like "what cancels what." Also note, the "per" helps indicate the relationship that's implied by the "conversion factor" in a way that the word "over" doesn't really help with.

How we read it:

One-half of a mole of nitrogen gas and one-half of a mole of oxygen gas and ninety point two nine kilojoules of energy yields one mole of nitrogen oxide gas.

Reasoning:

We use moles because it really is moles.

How we read it:

Carbon as graphite and oxygen gas yields carbon dioxide gas, with a change in enthalpy of negative 393.5 kilojoules. Carbon dioxide gas yields diamond and oxygen gas, with a change in enthalpy of positive 395.4 kilojoules. When these two equations are added together, oxygen gas and carbon dioxide gas cancel . The result is graphite yields diamond with a change in enthalpy of positive 1.9 kilojoules.

How we read it:

R equals k times A in brackets to the n power times B in brackets to the m power.

where R represents the reaction rate, k is the specific rate constant and A in brackets and B in brackets represent the molar concentrations of reactants A and B....

...

...replace the proportionality symbol with the constant k to write the rate law:

The rate is equal to the rate constant times the concentration of hydrogen times the square of the concentration of the nitrogen oxide.

Reasoning:

In this case, there is value to letting students know what the notation looks like. Because the generic equation is immediately followed by the spelled-out-explanation of what the symbols in the equation mean, that's another reason why we suggest "read symbol-for-symbol." But at the end, the specific example is not explained, so we speak in plain English.

How we read it:

The equilibrium constant is equal to the square of the concentration of hydrogen iodide, divided by the concentration of hydrogen gas, and divided by the concentration of iodide gas.

Reasoning:

Although the second "divided by" is repetitious and not strictly necessary, it helps the intelligibility of it.

How we read it:

Aqueous H A and liquid H 2 O is in equilibrium with aqueous H 3 O plus and aqueous A minus.

The equilibrium constant is equal to the concentration of H 3 O plus times something you select: Is it the concentration of H 2 O or the concentration of A minus?

How we read it:

Aqueous hydronium ions and aqueous ammonia are in equilibrium with aqueous ammonium ions and water.

One point six times ten to the negative 3 grams of calcium fluoride per 100 grams of water times one gram of water per one milliliter of water, times ten to the third milliliters per one liter, times one mole of calcium fluoride per seventy eight point one grams of calcium fluoride equals two point zero five times ten to the minus four moles per liter of calcium fluoride.

How we read it:

A unit of a pink complex of aqueous cobalt two plus ions containing six units of water and four units of aqueous chloride ions are in equilibrium with a unit of a blue complex of aqueous cobalt two plus ion containing four units of chloride ions and six water molecules.

How we read it:

The reaction equation for the cobalt(II) chloride equilibrium system is

A unit of aqueous complex of cobalt two plus ions combined with six water molecules and four units of aqueous chloride ions are in equilibrium with a unit of aqueous complex of cobalt two plus ions combined with four chloride ions and six units of liquid water.

The equilibrium expression for this reaction is written as:

K e q equals the concentration of the complex of cobalt two plus ion combined with four units of chloride ions, divided by the concentration of the complex of cobalt two plus ion combined with six units of water, divided by the concentration of chloride ion to the fourth power.