The bromate ion (by analogy with the chlorate ion ClO₃^-) is BrO₃^-, and we are told the product is Br₂. This is the basis of the half-reaction we need to balance.

This is the balanced half-reaction with ten electrons on the left.

Potassium must have 19 protons since its atomic number is 19. It must have 21 neutrons so the mass can add up to 40 as indicated by its name, potassium-40. Finally, in compounds potassium has a +1 charge, meaning it has one less electron than the number of protons, or 18.

For an isothermal expansion the temperature does not change, and therefore the average kinetic energy stays the same. Because the gas is ideal, there is no change in potential energy. That means that ΔE = 0 = w + q. Therefore q = -w, and the correct response is (D).

Half-reactions 1 and 2 are reductions because the charge of the copper ions decreases in the process. The iron half-reaction is an oxidation because the oxidation number increases from +2 to +3.

The difference in electronegativity is 1.9. The others are (a) 1.1, (c) 0.5, and (d) 0.6. This answer can be estimated since the P and F are more widely separated than the other pairs that are adjacent atoms in the periodic table.

The square pyramid is the shape of the molecule considering only the location of the atoms.

Mathematically, the fraction left is 1 microgram/10 kilograms, which is a ratio of (10^-6)/(10 × 10³) or 1/10^-10. For half-lives, the fraction remaining is (1/2)ⁿ, where n is the number of half-lives. By equating (1/2)ⁿ = (1/10^-10), we can take the logarithm of both sides of the equation to get n log (1/2) = -10. We can invert the fraction within the log term to get n log (2) = 10. (Note that this operation changes the sign.) Finally, n = 10/(log 2) = 33.2. This needs to be rounded to the next highest half-life, 34. This may be solved without a calculator by recalling that log (2) = 0.3.

This is a strong acid reacting with a strong base. We need to find out how much of either the strong acid or the strong base is left after the reaction is complete. This is a type of limiting-reactant problem. The reaction is HBr + KOH → KBr + H₂O.

With solutions it is easiest to calculate the moles of each reactant as

Since 1 mole of HBr reacts with 1 mole of KOH, we can see that 1.15 × 10^-4 mol of each will react. We can also see that ALL of the KOH is used up and must be the limiting reactant. The HBr is the excess reactant, and (1.25 × 10^-4 - 1.15 × 10^-4) = 1.0 × 10^-5 mol of HBr is left over. The molarity of HBr is the moles left over divided by the liters of solution, or

Since HBr is a strong acid, it dissociates completely, and [H⁺] = 0.00010 M

The pH = –log (0.00010) = 4.00.

The definition of molality is moles of solute per kilogram of solvent. In dilute solutions the mass of solute is negligible and the mass of one liter of water is one kilogram. Therefore, molarity and molality are essentially equal. We will calculate molarity as (g/M)/(L of solution)

At high temperature the molecules have the largest kinetic energy and pass each other rapidly to minimize interactions. The low pressure means few molecules in a given volume that increases the distance between molecules and decreases interactions. Response (C) is attractive, but high volume does not guarantee low pressure and minimal interactions.

The symbol for antimony is Sb. W is tungsten. Fe is iron. An and At do not exist.

NaC₂H₃O₃ → Na⁺ + C₂H₃O₂^-

The ethanoate ion then hydrolyzes water in the reaction

C₂H₃O₂^- + H₂O → HC₂H₃O₂ + OH^-

The molarity of the salt is 0.300 mol/0.500 L = 0.600 mol/L.

Assume that 0.600 >> x, so that 0.600 - x = 0.600

Silicon is a metalloid and is better known as a semiconductor material used in transistors.

This requires electrons to fill each available orbital in an energy level before pairing. Response (B) is a form of the Pauli exclusion principle. Responses (A) and (C) have little meaning.

The sum of the coefficients is 16. The reaction is balanced in steps. First write two half-reactions:

Balance all atoms except H and O:

Balance oxygens by adding water molecules:

Balance hydrogens by adding H⁺:

Balance charges with electrons:

Multiply the second half-reaction by 5 so both equations have the same number of electrons (now the e^- will cancel):

Add the two equations:

Cancel like items for the final equation:

Remember, I₂ and H₂O have coefficients of 1 although they are not explicitly written.

The reaction is Br₂(g) + Cl₂(g) → 2 BrCl(g)

The equilibrium expression is 

The initial concentration of BrCl is = (10.0 g/115 g mol^-1)/15 L = 5.8 × 10^-3 mol/L.

If 2x moles of BrCl reacts, x moles each of Br₂ and Cl₂ will be formed. (You may want to set up an equilibrium table at this point.)

Substituting into the equilibrium expression gives 

Take the square root of the entire equation:

The percentage of chlorine in the sample is calculated as

We know the grams of sample = 2.35 grams. We need to calculate the grams of chlorine from the 0.435 gram of AgCl precipitated. We start with ? g Cl = 0.435 g AgCl and add the conversion factors shown below.

Substitute the data into the above equation to get

The nitrogen on NH₄⁺ has an octet of electrons. Boron forms compounds with only six electrons, and the remaining compounds are in period 3 or above and may utilize d orbitals to expand the octet on the central atom to more than eight electrons. The others all have expanded octets.

The  quantum number designates the shape of an orbital, n designates the distance from the nucleus, and m designates the arbitrary direction in space of the orbital.

The conjugate acid of a base is obtained by adding one hydrogen atom and one positive charge to the base. Therefore, H₂PO₄^-+ H⁺ → H₃PO₄.

The ideal solution has the same attractive forces, or potential energy, in the mixture as in the pure solvents. This also means that mixing is neither exothermic nor endothermic.

The transition state is part of the transition state theory, not the collision theory.

All of the compounds are written correctly, and all the elements have the same number of atoms on each side of the arrow. Na is not balanced in response (A), Ag is not balanced in response (B), and the K is not balanced in response (D).

Taking 100 g of solution, 50 g will be NaOH and 50 g will be H₂O. The moles of each is

Rate laws cannot be determined from the reaction stoichiometry. Experimental data is always necessary to determine the rate law. One exception is a reaction that is an elementary step in a mechanism.

We rearrange the ideal gas law PV = nRT to PV = (g/M)RT

Dividing by PV and multiplying by the molar mass gives:

M = (g/V)(RT/P)

We will use R = 0.0821 L atm mol^-1.

Therefore we need to convert 722 torr to atm:

Entering the data and solving, we get

The nitrate ion has two oxygen atoms bonded to the nitrogen with single bonds, and one oxygen is bonded to the nitrogen with a double bond. This adds up to three sigma and one pi bond.

The question put into mathematical terms is

? mL KOH = 50.0 mL H₃PO₄

A balanced chemical equation is needed. The complete neutralization of H₃PO₄ is

H₃PO₄ + 3 KOH → K₃PO₄ + 3 H₂O

We change molarity units to mol/L or mol/1000 mL to use them as conversion factors:

Only reactions (C) and (D) are balanced. Only reaction (C) simply subtracts -2.37 from +0.80 to get the correct +3.17 V.