Online Chemistry 101 Exam
-
A particular chemical reaction involves a single reactant. What is the order of the reaction if the rate increases by a factor of eight when the concentration of the reactant is doubled?
-
0th order
-
1st order
-
2nd order
-
3rdorder
-
About This Quiz
This online chemistry 101 exam assesses knowledge on solutions and solubility, including calculations of molality and percent composition, understanding of solubility factors, and specific properties of solutions like density and concentration. It's designed to test essential chemistry skills relevant for academic progression in the field.
Quiz Preview
- 2.
Molality is :
-
Moles of solute / moles solvent
-
Moles solute / Kg solution
-
Moles solute / liters solution
-
Moles solute / Kg solvent
Correct Answer
A. Moles solute / Kg solventExplanation
Molality is a measure of the concentration of a solute in a solution. It is defined as the moles of solute divided by the kilograms of solvent. This is different from molarity, which is the moles of solute divided by the liters of solution. Therefore, the correct answer is "moles solute / Kg solvent".Rate this question:
-
- 3.
Which of the following would be expected to affect the rate of a given chemical reaction? I. The reaction temperature. II. The concentration of the reactants. III. A catalyst
-
I
-
II
-
I and III only
-
I,II and III
Correct Answer
A. I,II and IIIExplanation
The rate of a chemical reaction can be affected by several factors. Increasing the reaction temperature generally increases the rate of the reaction as it provides more energy for the reactant particles to collide and react. The concentration of the reactants also affects the rate of the reaction. Increasing the concentration increases the number of reactant particles, leading to more frequent collisions and a higher reaction rate. A catalyst can also affect the rate of a chemical reaction by providing an alternative pathway with lower activation energy, which allows the reaction to proceed faster. Therefore, all three factors, I, II, and III, would be expected to affect the rate of a given chemical reaction.Rate this question:
-
- 4.
What is the half-life for the reaction assuming first-order kinetics if 75% of a reactant decomposes in 60 minutes?
-
120 min
-
15 min
-
90 min
-
30 min
Correct Answer
A. 30 minExplanation
In a first-order reaction, the half-life is the time required for half of the reactant to decompose. Since 75% of the reactant decomposes in 60 minutes, it means that only 25% of the reactant remains after 60 minutes. Therefore, it takes an additional 30 minutes (half of 60 minutes) for the remaining 25% to decompose, resulting in a total of 90 minutes for the reactant to decompose by 50%. Hence, the correct answer is 30 minutes.Rate this question:
-
- 5.
The solubility of a gas in a liquid depends on ________
-
Tempreture
-
Pressure
-
Nature of the gas
-
All of the above
Correct Answer
A. All of the aboveExplanation
The solubility of a gas in a liquid depends on temperature, pressure, and the nature of the gas. Temperature affects the solubility because as temperature increases, the kinetic energy of the gas molecules also increases, leading to a decrease in solubility. Pressure also affects solubility because an increase in pressure increases the number of gas molecules in contact with the liquid, thereby increasing solubility. Additionally, the nature of the gas plays a role as different gases have different solubilities in a given liquid. Therefore, all of these factors contribute to the solubility of a gas in a liquid.Rate this question:
-
- 6.
Calculate the molrity of phosphoric acid in a solution that is 84% phosphoric acid and has a density of 1.87 g/ml
-
12
-
14
-
16
-
18
Correct Answer
A. 16Explanation
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution in liters. In this case, we need to determine the moles of phosphoric acid in the solution. Since we are given the density of the solution, we can calculate the volume of the solution using the formula density = mass/volume. Rearranging the formula, we find that volume = mass/density. Multiplying the volume by the percentage of phosphoric acid (0.84), we get the mass of phosphoric acid in the solution. Finally, dividing the mass by the molar mass of phosphoric acid (97.99 g/mol), we can calculate the moles of phosphoric acid. Dividing the moles by the volume of the solution in liters will give us the molarity.Rate this question:
-
- 7.
Which one of the following varies with tempreturre
-
Molarity
-
Mass percent
-
Mole fraction
-
Molality
Correct Answer
A. MolarityExplanation
Molarity is a measure of the concentration of a solution, defined as the number of moles of solute per liter of solution. As temperature increases, the volume of a solution usually expands due to thermal expansion. Since molarity is dependent on the volume of the solution, it will also change with temperature. Therefore, molarity is the only option given that varies with temperature.Rate this question:
-
- 8.
Gaseous N2O5 decomposes according to the following equation: N2O5(g) -------> NO2(g) + 1/2 O2(g) The experimental rate law is -Δ[N2O5]/Δt = k[N2O5]. At a certain temperature the rate constant is k = 5.0 x 10-4/second. In how many seconds will the concentration of N2O5 decrease to one-tenth of its initial value?
-
2.0 x 103 s
-
4.6 x 103 s
-
2.1 x 102 s
-
1.4 x 103 s
Correct Answer
A. 4.6 x 103 sExplanation
The rate law for the decomposition of N2O5 is given as -Δ[N2O5]/Δt = k[N2O5]. This means that the rate of the reaction is directly proportional to the concentration of N2O5. In this case, we are given that the rate constant k is 5.0 x 10-4/second.
To find the time it takes for the concentration of N2O5 to decrease to one-tenth of its initial value, we can use the integrated rate law for a first-order reaction. The integrated rate law for a first-order reaction is ln([N2O5]t/[N2O5]0) = -kt, where [N2O5]t is the concentration of N2O5 at time t and [N2O5]0 is the initial concentration.
In this case, we want to find the time it takes for [N2O5]t to be one-tenth of [N2O5]0. So we can rewrite the equation as ln(1/10) = -kt. Rearranging the equation, we get t = ln(1/10)/k.
Plugging in the value of k, we get t = ln(1/10)/(5.0 x 10-4/second) ≈ 4.6 x 103 seconds. Therefore, the concentration of N2O5 will decrease to one-tenth of its initial value in approximately 4.6 x 103 seconds.Rate this question:
-
- 9.
At 350 K, a particular second-order reaction, consisting of a single reactant, A, has a rate constant equal to 4.5 x 10-3 s-1. If the initial concentration of A is 0.80 M, how many half-lives are required for the concentration of A to become equal to 0.10 M?
-
1
-
2
-
3
-
4
Correct Answer
A. 3Explanation
At 350 K, the rate constant for the second-order reaction is given as 4.5 x 10-3 s-1. The reaction follows the equation: A → products. The initial concentration of A is 0.80 M, and we want to find the number of half-lives required for the concentration of A to become 0.10 M.
In a second-order reaction, the half-life is inversely proportional to the initial concentration. As the initial concentration decreases, the number of half-lives required to reach a certain concentration increases.
To find the number of half-lives, we can use the formula: t1/2 = 1 / (k * [A]0), where t1/2 is the half-life, k is the rate constant, and [A]0 is the initial concentration.
Plugging in the values, we get: t1/2 = 1 / (4.5 x 10-3 s-1 * 0.80 M) = 3.47 x 102 s.
Now, we can find the number of half-lives required by dividing the total time (t) by the half-life (t1/2): t / t1/2 = (ln([A]0 / [A])) / (ln(2)).
Plugging in the values, we get: (ln(0.80 M / 0.10 M)) / (ln(2)) = 2.08.
Since we need a whole number of half-lives, we round up to the nearest whole number, which is 3. Therefore, the answer is 3.Rate this question:
-
- 10.
Consider the following chemical reaction: 2 H2O2(aq) ------> 2 H2O(l) + O2(g) If the average rate of disappearance of H2O2 over a certain time interval is 6.80 x 10-5 M s-1 what is the average rate of appearance of O2 during this same time interval?
-
4.62 x 10-9 M s-1
-
3.40 x 10-5 M s-1
-
6.80 x 10-5 M s-1
-
1.36 x 10-4 M s-1
Correct Answer
A. 3.40 x 10-5 M s-1Explanation
In the given chemical reaction, 2 moles of H2O2 produce 1 mole of O2. Therefore, the stoichiometric ratio between the disappearance of H2O2 and the appearance of O2 is 2:1. Since the average rate of disappearance of H2O2 is given as 6.80 x 10-5 M s-1, the average rate of appearance of O2 would be half of that, which is 3.40 x 10-5 M s-1.Rate this question:
-
- 11.
Which of the following would DECREASE the rate of a chemical reaction?
-
Decreasing the activation energy.
-
Increasing the concentrations of the reactants.
-
Increasing the temperature.
-
Adding a catalyst.
-
None of these will decrease the rate.
Correct Answer
A. None of these will decrease the rate.Explanation
Increasing the concentrations of the reactants, increasing the temperature, and adding a catalyst are all factors that would typically increase the rate of a chemical reaction. However, the given answer states that none of these options will decrease the rate. This means that none of the options provided would have a negative effect on the rate of the reaction.Rate this question:
-
- 12.
Which of the following very high solubility in water
-
C6H6
-
C2H5OH
-
C6H5NH2
-
C6H5OH
Correct Answer
A. C2H5OHExplanation
C2H5OH (ethanol) has very high solubility in water due to its ability to form hydrogen bonds with water molecules. Ethanol contains a hydroxyl group (OH), which can form hydrogen bonds with the water molecules. This allows ethanol to dissolve easily in water, making it highly soluble. On the other hand, C6H6 (benzene) and C6H5NH2 (aniline) do not have the ability to form hydrogen bonds with water, so their solubility in water is significantly lower.Rate this question:
-
- 13.
A solution is said to contain 28% phosphoric acid by mass . what does this mean ?
-
1 ml of this solution contains 28 g of phosphoric acid
-
1 L of this solution has a mass of 28 g
-
10 g of this solution contains 28 g of phosphoric acid
-
The density of this solution is 2.8 g/ml
Correct Answer
A. 10 g of this solution contains 28 g of phosphoric acidExplanation
This means that for every 10 grams of this solution, there are 28 grams of phosphoric acid.Rate this question:
-
- 14.
A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water the density of the resulting solution is 1.05 g/ml . calculate the % by weight of CaCl2 in the solution described above
-
5.94%
-
6.32%
-
0.0632%
-
0.0594%
Correct Answer
A. 5.94%Explanation
The % by weight of CaCl2 in the solution can be calculated by dividing the mass of CaCl2 by the total mass of the solution and multiplying by 100. The mass of CaCl2 can be calculated by converting the given mass of CaCl2 into grams and subtracting it from the total mass of the solution. The total mass of the solution can be calculated by multiplying the density of the solution by the volume of the solution. After performing the calculations, it is found that the % by weight of CaCl2 in the solution is 5.94%.Rate this question:
-
- 15.
A student determined the value of the rate constant, k, for a chemical reaction at several different temperatures. Which of the following graphs of the student's data would give a straight line?
-
K versus T
-
K versus (1/T)
-
Ln k versus (1/T)
-
Ln k versus T
Correct Answer
A. Ln k versus (1/T)Explanation
The graph of ln k versus (1/T) would give a straight line. This is because the Arrhenius equation, which relates the rate constant (k) to temperature (T), can be rearranged to give ln k as a linear function of 1/T. Therefore, plotting ln k against (1/T) would result in a straight line, where the slope of the line can be used to determine the activation energy of the reaction.Rate this question:
-
Quiz Review Timeline (Updated): Mar 20, 2023 +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
-
Current Version
-
Mar 20, 2023Quiz Edited by
ProProfs Editorial Team -
Mar 27, 2018Quiz Created by
Mohammad_aa1997
Back to top