For this reaction, there is no temperature from absolute zero on up where the ΔG° will not be negative.

The following equation is needed for this question:

C4H10 + O2 -> H2O + CO2

To balance this equation, a 2 should be placed in front of C4H10 to make it easier to balance.
2 C4H10 + O2 -> H2O + CO2

Next, an 8 must be placed on the right side in front of CO2 to balance the carbons.
2 C4H10 + O2 -> H2O + 8 CO2

Next, a 10 must be placed on the right side, in front of H2O to balance the hydrogens.
2 C4H10 + O2 -> 10 H2O + 8 CO2

Finally, a 13 must be placed on the left side, in front of O2 to balance oxygen.
2 C4H10 + 13 O2 -> 10 H2O + 8 CO2

Thus, the coefficients needed to balance this equation are: 2, 13, 10, 8, or answer option [E].

We need to write and balance the equation first:

2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

Start with the molarity of H₂SO₄. Use dimensional analysis to convert it to the molarity of Al(OH)₃ as follows:

When we extract the coefficients from the right side of this equation, it matches response (A).

Butan-1-ol (butanol), CH₃CH₂CH₂CH₂OH, can hydrogen-bond, whereas the other compounds cannot. London forces (instantaneous dipoles) are also present, but each compound is roughly the same length, and these forces will be similar for all five molecules.

The ideal gas law, PV = nRT when divided by P results in: (nRT) / P = V, assuming that there is 1 mol at STP, this would equal 22.4L, which is also known as the molar volume of an ideal gas.

Beaker: holds liquids, should not be used for accurate measurement.

Buret: accurately measures unknown volumes up to 50.00 mL.

Volumetric Pipette: accurately measures known small volumes like; 5.00 ml, 10.00 mL, 20.00 mL, 25.00 mL and 50.00 mL.

Measuring cylinder: good for measuring volumes up to a 2 L, but not as accurately.

Volumetric flask: specifically prepared to measure larger known volumes with an accuracy of 1 mL in a Liter.

All options are common laboratory errors, but Parallax Error may occur when a pointer on a meter or surface of a liquid in a buret or pipette is not observed from directly straight on. If they are observed from above or below then the liquid surface or pointer will not be seen to line up with the scale.

TNT explosions are very fast reactions (the lower the activation energy the less a reaction is impeded) involving the production of enormous quantities of heat (exothermic) and gaseous products (the production of hot gases from a solid explosive dramatically increases the disorder of the system, thus high entropy).

All four responses are correct applications of E°_cell.

The dissolution reaction is Fe(OH)₃(s)  Fe³⁺ + 3 OH^-, and the K_sp equation is K_sp = [Fe³⁺][OH^-]³.

If s mol/L of Fe(OH)₃ dissolves, the solution will contain s mol/L of Fe³⁺ and 3s mol/L of OH^-.

Substituting into the K_sp equation results in

(the fourth root can be obtained by taking successive square roots or by raising the number to the 1/4 or 0.25 power using the y^x function found on mos

Conjugate acid base pairs always differ by one H⁺. Only the ions in (C) meet this criterion.

We calculate the grams of ethanoic acid needed to react with the given amount of ethyl alcohol. Two important conversion factors are

1 mol CH₃COOH = 60 g CH₃COOH

and

1 mol CH₃CH₂OH = 44 g CH₃CH₂OH

Setup: ? g CH₃COOH = 100 g CH₃CH₂OH

Applying the above conversion factors we get

The problem only gave us 100 g CH₃COOH. Therefore CH₃COOH is the limiting reactant. We now calculate the mass of ethyl ethanoate formed from the GIVEN mass of 100 g CH₃COOH as follows.

Set up the question.

? g CH₃CH₂OC=OCH₃ = 100 g CH₃COOH

Apply conversion factors

NOTE: We get the same result if we convert each reactant to grams of ethyl ethanoate and then choose the smaller of the two results.

To determine the empirical formula we must calculate the simplest ratio of moles of each element in the formula. The percentages are converted to grams by assuming 100 g of sample and changing all percentages to grams. The moles are then calculated as

Divide each of the answers above by 0.541 to get 7.98 mol C, 8.98 mol H, and 1.00 mol Br. The empirical formula is C₈H₉Br.

All calcium atoms contain 20 protons and 20 electrons. Depending on the isotope, a calcium atom may or may not have 20 neutrons. Finally, the mass in the periodic table is a weighted average of isotopes and is NOT the mass of any calcium atom.

Density is calculated by dividing the grams of a substance by the volume it displaces, in either mL or L. In this case the density would be calculated as: 7.000g / 21.0mL = 0.333 g/mL

Note: The correct answer must have 3 significant digits because the 21.0 mL number used in the calculation had 3 significant digits. The rule for multiplying and dividing significant digits is the answer must have the same number of significant digits as the number with the least amount of significant digits.

The temperature (in Kelvins) of a gas is directly proportional to the average kinetic energy of the gas. Therefore, if the temperature is quadrupled, so to is the average kinetic energy of the gas.

It is risky to say that all compounds have any given property. However, given the solubility rules generally used in chemistry courses, ammonium ions for the most part indicate a soluble ionic substance. The other items listed all have well-known insoluble substances.

Dashed lines traditionally represent attractions that are NOT covalent bonds. In this case, they represent hydrogen bonds.

We are asked for the theoretical yield of iron. Usually, the theoretical yield is expressed in units of grams of product, as all the responses imply. This is a limiting-reactant problem because the mass of both reactants is given. We will solve it by calculating the mass of iron that can be made from 2.00 g C assuming that Fe₂O₃ is the excess reactant. Then we will calculate the mass of iron that we can prepare from 26.0 g Fe₂O₃ assuming that carbon is the excess reactant. We set up the two equations as

Carbon produces the smaller amount of iron. Therefore, carbon is the limiting reactant, and 12.4 g of iron is the theoretical yield.

All the other statements violate the law of conservation of energy.

A pressure cooker fixes the atmospheric pressure to be higher than normal. In order for a liquid to boil, the vapor pressure inside of the container must be at least equal to the atmospheric pressure. For higher atmospheric pressures, like in a pressure cooker, the vapor pressure must also increase. Therefore, it takes longer for the vapor pressure of the gas to accumulate, leaving the liquid to become hotter and hotter. Once the boiling point is reached, the liquid is hotter than normal, cooking the food faster. Once the vapor pressure equals the environmental pressure of the pressure cooker, it remains constant. At this point the liquid is boiling and the vapor pressure of the liquid is equal to the atmospheric pressure of the pressure cooker. In this case, the only answer option that aligns correctly to what occurs in a pressure cooker is [D].

The rate-determining step governs the overall reaction rate. Proposed mechanisms must provide appropriate steps to include the slow step.

Response (C) is the only one that has different phases noted.

The chemical equation tells us nothing about the position of equilibrium or if the reaction is thermodynamically favored.

The answer options are all nonpolar compounds. Non-polar compounds have only London dispersion as their intermolecular force, which is relatively weak in comparison to the other intermolecular forces. However, the strength of London Dispersion increases with higher molecular mass. Therefore, the compound with the highest boiling point would be the compound with the strongest intermolecular force, which in this case is only London dispersion. Answer option [E] would have the highest boiling point of the answer options as it has the highest molecular mass.

We can use the approximation that the square root of the K_b times the initial concentration will give us the [OH^-]. Taking the square root of 4.4 × 10^-6 yields approximately 2 × 10^-3. So the pOH will be a little less than 3. Converting to pH results in a range of 10 to 12.

We calculate that the mass of Mg needed to produce 50 mL of H₂ is approximately 0.05 g, so 0.1 g of water left will be a 200 percent error. Recording 25 °C rather than 298K is a 1000 percent error. Since the total gas volume is 50 mL, ±20 mL is a 40 percent error. The vapor pressure error is less than 5 percent and is the smallest error.