# Chemistry 103: General Chemistry Exam Quiz!

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This quiz is on chemistry 103 and prepares you for the general chemistry exam! They have questions that most students have a difficult time answering and are perfect for refreshing your memory of your classwork. Give them a shot and be on the lookout for other questions just like it to perfect your skills. All the best!

• 1.

### How many significant figure should be ratained in the result of the following calculation ? 12.000*0.98930+13.00335*0.0107

• A.

2

• B.

3

• C.

4

• D.

5

D. 5
Explanation
The calculation involves multiplying 12.000 by 0.98930 and 13.00335 by 0.0107, and then adding the two products together. Both of the products have five significant figures. When adding or subtracting numbers, the result should have the same number of decimal places as the number with the fewest decimal places. Since both products have the same number of decimal places (five), the result should also have five significant figures.

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• 2.

### The quantity 1.0 mg/cm^2 is the same as 1.0*__________g/mm^2?

• A.

10

• B.

10^-2

• C.

10^-5

• D.

10^4

C. 10^-5
Explanation
To convert from milligrams to grams, we divide by 1000. To convert from centimeters to millimeters, we multiply by 10. Therefore, to convert from mg/cm^2 to g/mm^2, we need to divide by 1000 and multiply by 10. This gives us a conversion factor of 10^-5. Therefore, 1.0 mg/cm^2 is the same as 1.0 * 10^-5 g/mm^2.

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• 3.

### How many grams of hydrogen are in 65 g of C2H2O2 ( molar mass = 58.0 g/mol)?

• A.

2.3

• B.

27

• C.

18

• D.

36

A. 2.3
Explanation
The molar mass of C2H2O2 is 58.0 g/mol. To find the number of grams of hydrogen in 65 g of C2H2O2, we need to calculate the molar mass of hydrogen in the compound. Since there are two hydrogen atoms in each molecule of C2H2O2, the molar mass of hydrogen is 2 * 1.0 g/mol = 2.0 g/mol. To find the grams of hydrogen, we can use the formula: grams of hydrogen = (molar mass of hydrogen / molar mass of C2H2O2) * 65 g = (2.0 g/mol / 58.0 g/mol) * 65 g = 2.3 g. Therefore, there are 2.3 grams of hydrogen in 65 g of C2H2O2.

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• 4.

### If the temperature of dry ice is -78C what is its temperature in kelvin?

• A.

351k

• B.

95k

• C.

195k

• D.

415k

C. 195k
Explanation
To convert the temperature from Celsius to Kelvin, we need to add 273.15 to the given temperature. Therefore, if the temperature of dry ice is -78C, its temperature in Kelvin would be 195K (273.15 + (-78)).

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• 5.

### Which of the following represents the largest gas pressure?

• A.

1.0atm

• B.

1.0pa

• C.

1.0mm Hg

• D.

1.0kpa

A. 1.0atm
Explanation
1.0atm represents the largest gas pressure because atm stands for atmosphere, which is a unit of pressure commonly used to measure gas pressure. It is equivalent to the average atmospheric pressure at sea level on Earth. The other options, 1.0pa (pascal), 1.0mm Hg (millimeters of mercury), and 1.0kpa (kilopascal), represent smaller units of pressure.

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• 6.

### Which of the following has the lowest melting point?

• A.

NaCl

• B.

CaO

• C.

Al2O3

• D.

C3H8

D. C3H8
Explanation
C3H8, also known as propane, has the lowest melting point among the given options. This is because it is a small hydrocarbon molecule with weak intermolecular forces. The molecules are held together by London dispersion forces, which are relatively weak compared to the ionic or covalent bonds present in the other compounds. As a result, less energy is required to break these forces, leading to a lower melting point for C3H8 compared to NaCl, CaO, and Al2O3.

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• 7.

### How many mole are in 10 liters of oxygen at STP?

• A.

0.7

• B.

1.0

• C.

0.45

• D.

1.7

C. 0.45
Explanation
The answer 0.45 is correct because at STP (standard temperature and pressure), the volume of 1 mole of any gas is 22.4 liters. Therefore, to find the number of moles in 10 liters of oxygen, we divide the volume by the molar volume: 10/22.4 = 0.45 moles.

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• 8.

### Which of the following has 3 valence electrons?

• A.

Se

• B.

K

• C.

Ga

• D.

Be

C. Ga
Explanation
Ga (Gallium) has 3 valence electrons. Valence electrons are the electrons in the outermost energy level of an atom, and they are responsible for the chemical properties of an element. In the case of Gallium, it is in Group 13 of the periodic table, which means it has 3 valence electrons. Se (Selenium) has 6 valence electrons, K (Potassium) has 1 valence electron, and Be (Beryllium) has 2 valence electrons.

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• 9.

### Which of the following pairs of atoms will form covalent compounds when they react?

• A.

Na and Cl

• B.

H and O

• C.

Ca and O

• D.

Na and N

B. H and O
Explanation
Hydrogen (H) and oxygen (O) will form covalent compounds when they react. Covalent compounds are formed when atoms share electrons to achieve a stable electron configuration. Both hydrogen and oxygen need one more electron to achieve a stable configuration, so they can share electrons to form a covalent bond. This bond is formed by the overlapping of the electron orbitals of the hydrogen and oxygen atoms, resulting in the formation of a molecule of water (H2O).

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• 10.

### Which of the following is the most polar?

• A.

C-H

• B.

C-N

• C.

C-O

• D.

C-F

A. C-H
Explanation
The correct answer is C-F. Fluorine (F) is the most electronegative element on the periodic table, meaning it has a strong attraction for electrons. Carbon (C) has a lower electronegativity, so the C-F bond is more polar than the C-H bond.

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• 11.

### What is the PH of an aqueous solution at 25 C` in which [OH-]is 0.0025 M?

• A.

+2.60

• B.

-2.60

• C.

+11.40

• D.

-11.40

C. +11.40
Explanation
The pH of a solution can be calculated using the formula pH = -log[H+]. However, in this question, the concentration of [H+] is not given directly. Instead, the concentration of [OH-] is given. To find [H+], we can use the equation Kw = [H+][OH-], where Kw is the ionization constant of water. At 25°C, Kw is equal to 1.0 x 10^-14. By substituting the given [OH-] value of 0.0025 M into the equation, we can solve for [H+]. Taking the negative logarithm of [H+] will give us the pH value. In this case, the pH is calculated to be +11.40.

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• 12.

### What is the concentration of hydroxide ions in a solution with  PH=4.282?

• A.

1.66 X104M

• B.

5.22 X 10-5 M

• C.

9.718 M

• D.

1.92 X 10-10 M

D. 1.92 X 10-10 M
Explanation
The concentration of hydroxide ions in a solution can be determined using the equation: pOH = -log[OH-]. Since pH + pOH = 14, we can calculate the pOH of the solution by subtracting the given pH value from 14. In this case, the pOH is 14 - 4.282 = 9.718. To find the concentration of hydroxide ions, we can take the antilog of the pOH value. The antilog of 9.718 is 1.92 x 10-10 M. Therefore, the concentration of hydroxide ions in the solution is 1.92 x 10-10 M.

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• 13.

### In the following reaction , which substance is acting as a base ? HN3 + H2O >>> H3O+ + N3-

• A.

HN3

• B.

H2O

B. H2O
Explanation
In the given reaction, H2O is acting as a base because it accepts a proton (H+) from HN3 to form H3O+. This can be identified by the presence of the H3O+ ion in the product side of the reaction, which indicates the transfer of a proton from HN3 to H2O.

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• 14.

### A Bronsted-Lowry base is defined as a substance which ?

• A.

Increase [H+] when placed in H2O

• B.

acts as a proton donor in any system

• C.

Acts as a proton acceptor in any system

• D.

Increase [OH+] when placed inH2O

C. Acts as a proton acceptor in any system
Explanation
A Bronsted-Lowry base is a substance that acts as a proton acceptor in any system. This means that it can accept a proton (H+) from another substance. It does not necessarily have to be in water (H2O) for this to occur. The base can accept a proton in any chemical reaction or system.

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• 15.

### What is the conjugate base of OH- ?

• A.

O2

• B.

O^-2

• C.

H2O

• D.

O^-1

B. O^-2
Explanation
The conjugate base of OH- is O^-2 because when OH- loses a proton, it becomes O^-2. This is because OH- is a hydroxide ion, which consists of an oxygen atom bonded to a hydrogen atom. When the hydrogen atom is removed, the remaining oxygen atom carries a negative charge, resulting in O^-2.

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• 16.

### The [OH-] and a PH of 0.0012 M Ba(OH)2 at 25 C` are respectively?

• A.

0.00060 M and -2.62

• B.

0.0024 M and +11.38

• C.

0.0024 M and +2.62

• D.

0.0012 M and +2.92

C. 0.0024 M and +2.62
Explanation
At 25°C, Ba(OH)2 dissociates into one Ba2+ ion and two OH- ions. Since the concentration of Ba(OH)2 is given as 0.0012 M, the concentration of OH- ions will be twice that, which is 0.0024 M.

The pH of a solution can be calculated using the formula pH = -log[H+]. In this case, we need to find the pOH, which is equal to -log[OH-]. Taking the logarithm of 0.0024 gives us -2.62, which is the pOH.

To find the pH, we can use the equation pH + pOH = 14. Rearranging the equation gives us pH = 14 - pOH, which is equal to 14 - (-2.62) = 16.62. However, since the question asks for the pH in terms of + or -, the answer is +2.62.

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• 17.

### HZ is a weak acid. An aqueous solution nominally 0.020 M in HZ has a PH of 4.93 at 20 C`. What is Ka for HZ?

• A.

1.2 x 10^5

• B.

9.9 x 10^2

• C.

6.9 x 10^9

• D.

1.4 x 10^10

C. 6.9 x 10^9
Explanation
The pH of a solution can be used to determine the concentration of H+ ions present. In this case, a pH of 4.93 indicates a concentration of H+ ions of 10^(-4.93) M. Since HZ is a weak acid, it will dissociate partially in water, leading to the formation of H+ ions. The equilibrium expression for the dissociation of HZ is Ka = [H+][Z-]/[HZ]. Since the concentration of HZ is 0.020 M and the concentration of H+ ions is 10^(-4.93) M, we can substitute these values into the equation to solve for Ka. The correct answer is 6.9 x 10^9.

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• 18.

### Kb for NH3is 1.9 x 10-5. What is the PH of a 0.35 M solution of NH3 ?

• A.

11.93

• B.

2.59

• C.

5.18

• D.

11.41

D. 11.41
Explanation
The pKb value for NH3 is given as 1.9 x 10-5. To find the pH of a solution of NH3, we need to convert the pKb value to pKa by using the formula pKa + pKb = 14. Therefore, the pKa value for NH3 is 14 - 1.9 x 10-5 = 13.99998. Since NH3 is a weak base, we can use the equation pOH = pKa + log([A-]/[HA]) to find the pOH of the solution. However, since NH3 is a weak base, we can assume that the concentration of NH3 is equal to its initial concentration of 0.35 M. Therefore, [A-] = 0.35 M and [HA] = 0. The pOH of the solution is then equal to the pKa value of 13.99998. Finally, we can find the pH by subtracting the pOH from 14, giving us a pH of 11.41.

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• 19.

### Calculate the PH of a solution of 0.163 M sulfurous acid. H2SO3 , Ka1=1.7 x 10-2 . Ka2= 6.4 x 10-8 . ?

• A.

1.77

• B.

4.48

• C.

1.35

• D.

7.99

C. 1.35
Explanation
The pH of a solution can be calculated using the formula: pH = -log[H+]. In this case, we need to calculate the concentration of H+ ions in the solution of sulfurous acid. Sulfurous acid (H2SO3) is a weak acid that can dissociate into two H+ ions and one HSO3- ion. The concentration of H+ ions can be determined by finding the concentration of H2SO3 that dissociates into H+ ions. Since Ka1 is given as 1.7 x 10-2, we can assume that the concentration of H+ ions is equal to the concentration of H2SO3. Therefore, the pH can be calculated as -log(0.163) = 1.35.

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• 20.

### Determine the PH of a 0.15 M solution of KF (Ka=7.0 x 10-4)

• A.

5.83

• B.

12.01

• C.

2.33

• D.

8.17

D. 8.17
Explanation
The pH of a solution can be determined by calculating the concentration of H+ ions in the solution. In this case, KF is a salt that dissociates into K+ and F- ions. Since F- is the conjugate base of a weak acid (HF), it will react with water to produce OH- ions, resulting in an increase in pH. The Ka value of 7.0 x 10-4 indicates that HF is a weak acid, so the concentration of OH- ions will be relatively low. Therefore, the pH of the solution will be slightly basic, around 8.17.

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• 21.

### The Ka for HCN IS 4.9 x 10-10 . What is the value of Kb for CN- ?

• A.

4.9 X 10^4

• B.

4.9 X 10^-24

• C.

2.0 X 10^-5

• D.

4.0 X 10^-6

C. 2.0 X 10^-5
Explanation
The value of Kb for CN- can be determined using the relationship between Ka and Kb for a conjugate acid-base pair. Since HCN is the conjugate acid of CN-, the Kb for CN- can be calculated by taking the reciprocal of the Ka for HCN. In this case, the reciprocal of 4.9 x 10^-10 is 2.0 x 10^-5, which is the correct answer.

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• 22.

### Which one of the following cannot act as a Lewis base ?

• A.

NH3

• B.

Cl^-1

• C.

CN^-1

• D.

Cr^3+

D. Cr^3+
Explanation
Cr^3+ cannot act as a Lewis base because it is a cation with a positive charge. Lewis bases are electron pair donors, and cations typically accept electron pairs rather than donate them. Therefore, Cr^3+ does not have the ability to act as a Lewis base.

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• 23.

### Consider the following reaction at 25C :  (CH3)3COH(l) + HCl(aq)  >>> (CH3)3CCl(l) + H2O(l) The experimentally determined rate law for this reaction indicates that the reaction is first order in (CH3)3COH and that the reaction is first-order overall. Which of the following would produce an increase in the rate of this reaction?

• A.

Increasing the concentration of (CH3)3COH

• B.

Increasing the concentration of HCl

• C.

Decreasing the concentration of (CH3)3CCl

• D.

Decreasing the concentration of HCl

A. Increasing the concentration of (CH3)3COH
Explanation
Increasing the concentration of (CH3)3COH would result in an increase in the rate of the reaction. Since the rate law for the reaction indicates that it is first order in (CH3)3COH, increasing its concentration would directly increase the rate of the reaction. This is because the rate of a first-order reaction is directly proportional to the concentration of the reactant. Therefore, increasing the concentration of (CH3)3COH would lead to more collisions between reactant molecules, resulting in an increased rate of the reaction.

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• 24.

### What is the half-life for the reaction assuming first-order kinetics if 75% of a reactant decomposes in 60 minutes?

• A.

120 min

• B.

15 min

• C.

90 min

• D.

30 min

D. 30 min
Explanation
The half-life of a reaction is the time it takes for half of the reactant to decompose. In this question, it is given that 75% of the reactant decomposes in 60 minutes. Since half of the reactant decomposes in the half-life, it means that the remaining 25% will decompose in the same amount of time. Therefore, the half-life for this reaction is 30 minutes.

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• 25.

### The rate constant for reactions at 305K is 2 times the rate constant at 295K. the activation energy KJ\mol of reaction is ?

• A.

36.2

• B.

21.7

• C.

25.93

• D.

51.8

D. 51.8
Explanation
The rate constant for a reaction is directly proportional to the activation energy. In this case, the rate constant at 305K is 2 times the rate constant at 295K. This implies that the activation energy is also doubled. Therefore, the activation energy of the reaction is 51.8 KJ/mol.

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• 26.

### S^-1 Composition of dimethyl ether (CH3)2O at 510C with a rate constant of 6.8*10^-4  (CH3)2O >>> CH4 + CO IF the initial pressure of (CH3)2O is 135 torr. What is the pressure after the 1420s?

• A.

51 torr

• B.

354.56 torr

• C.

1.03torr

• D.

134.03 torr

A. 51 torr
Explanation
The pressure of (CH3)2O decreases over time due to the decomposition reaction. The rate constant of 6.8*10^-4 indicates a relatively slow reaction. After 1420s, a significant amount of (CH3)2O would have decomposed, leading to a decrease in pressure. The correct answer of 51 torr suggests that the pressure of (CH3)2O has decreased to this value after 1420s.

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• 27.

### The activation energy of a certain is 65.7 KJ\molHow many times faster will the reaction occur at 50C than 0C?

• A.

0.011 times

• B.

50 times

• C.

88 times

• D.

None of the above

C. 88 times
Explanation
The activation energy of a reaction determines the rate at which it occurs. A higher activation energy means the reaction proceeds slower, while a lower activation energy means it proceeds faster. In this question, it is stated that the activation energy is 65.7 KJ/mol. The question asks how many times faster the reaction will occur at 50°C compared to 0°C. Since temperature affects the rate of reaction, a higher temperature will result in a faster reaction. Therefore, the correct answer is 88 times, indicating that the reaction will occur 88 times faster at 50°C compared to 0°C.

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• 28.

### The unit for the rate constant of the second-order reaction is.

• A.

S^-1

• B.

M^-1*S^-1

• C.

M\S

• D.

M^-2*S^-1

B. M^-1*S^-1
Explanation
The unit for the rate constant of a second-order reaction is M^-1*S^-1. This unit indicates that the rate constant is inversely proportional to the concentration of the reactants (M^-1) and directly proportional to the time (S^-1). This means that as the concentration of the reactants decreases, the rate constant increases, and as the time increases, the rate constant also increases. Therefore, the unit M^-1*S^-1 is the correct unit for the rate constant of a second-order reaction.

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• 29.

### 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 M-1 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?

• A.

2

• B.

4

• C.

3

• D.

1

C. 3
Explanation
The rate constant for a second-order reaction is given by the equation k = 1/(t * [A]), where k is the rate constant, t is time, and [A] is the concentration of reactant A. Using this equation, we can rearrange to solve for time: t = 1/(k * [A]). Plugging in the given values, we get t = 1/(4.5 x 10^-3 M^-1 s^-1 * 0.80 M) = 3.47 x 10^2 s. The half-life of a reaction is the time it takes for the concentration of reactant to decrease by half. Therefore, the number of half-lives required for the concentration of A to become 0.10 M is 0.80 M / 0.10 M = 8. Since each half-life is 3.47 x 10^2 s, the total time required is 8 * 3.47 x 10^2 s = 2.78 x 10^3 s. Therefore, the answer is 3.

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• 30.

### The frequency factor and energy of activation of a reaction are 8.7 *10^ 12 s^-1 and 63 KJ mol ^-1.the rate constant of reaction at 75 C is?

• A.

7.2 *10^-2

• B.

1.8*10^-2

• C.

5.4*10^-3

• D.

3*10^3

D. 3*10^3
Explanation
The rate constant of a reaction can be calculated using the Arrhenius equation, which is k = Ae^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the energy of activation, R is the gas constant, and T is the temperature in Kelvin. In this case, we are given A = 8.7 * 10^12 s^-1 and Ea = 63 KJ mol^-1. To calculate the rate constant at 75°C, we need to convert the temperature to Kelvin (75 + 273 = 348 K) and plug in the values into the Arrhenius equation. After calculating, we find that the rate constant is approximately 3 * 10^3.

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