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What would happen if a scuba tank (a rigid, sealed gas cylinder) heats up?

a
The pressure exerted by the gas in the tank would increase.
b
The size of the gas particles in the tank would increase.
c
The number of particles in the tank would increase.
d
The density of the gas in the tank would increase.

11 answers

If a scuba tank (a rigid, sealed gas cylinder) heats up, the correct option would be:

a. The pressure exerted by the gas in the tank would increase.

This is because, according to Gay-Lussac's Law, if the volume of a gas is held constant (as it would be in a rigid tank), the pressure of the gas is directly proportional to its temperature in Kelvin. As the temperature increases, the kinetic energy of the gas molecules increases, which results in more frequent and forceful collisions with the walls of the tank, thus increasing the pressure.

The other options are incorrect for the following reasons:

  • b. The size of the gas particles in the tank would increase. Gas particles themselves do not change size with temperature. However, their kinetic energy increases, causing them to move more rapidly.

  • c. The number of particles in the tank would increase. The number of gas particles is determined by the amount of gas initially in the tank and does not change simply due to an increase in temperature.

  • d. The density of the gas in the tank would increase. Since the number of particles remains the same and the temperature increases (which typically leads to an increase in pressure), the density of the gas (mass/volume) does not increase; it may actually decrease if the gas expands under pressure. However, in a rigid tank, the volume doesn’t change, but density remains constant with a given amount of gas.

Thus, the most accurate answer is option a.

According to Newton's second law, what happens when the same force is applied to two objects of different masses?

a
The object with greater mass will experience a smaller acceleration.
b
The object with greater mass will experience a greater acceleration.
c
The object with smaller mass will experience a smaller acceleration.
d
The two objects experience the same acceleration.

According to Newton's second law of motion, which states that \( F = ma \) (where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration), the correct option in the scenario where the same force is applied to two objects of different masses would be:

a. The object with greater mass will experience a smaller acceleration.

This is because if you apply the same force to two objects, the object with the greater mass will have a smaller acceleration than the object with the smaller mass. The reason is that acceleration is inversely proportional to mass when the force is constant; hence:

\[ a = \frac{F}{m} \]

As mass \( m \) increases, for a constant force \( F \), the acceleration \( a \) decreases.

The other options are not correct for the following reasons:

  • b. The object with greater mass will experience a greater acceleration. This is incorrect because greater mass results in smaller acceleration under the same force.

  • c. The object with smaller mass will experience a smaller acceleration. This statement is misleading; while the object with the smaller mass experiences greater acceleration compared to the heavier object, it does not mean it experiences a smaller acceleration overall.

  • d. The two objects experience the same acceleration. This is incorrect, as they will not experience the same acceleration when the forces applied are the same and their masses differ.

Thus, the correct answer is a.

Match the description of nuclear decay to the type of nuclear decay.

1. 4/2 He

2. 0/-1 e

3. 0/0 y

can be stopped by paper

can be stopped by aluminum

can be stopped by lead

answers

gamma

alpha

beta

To match the types of nuclear decay with their descriptions and their ability to be stopped by different materials, we can use the information given:

  1. 4/2 He - This represents an alpha particle.

    • Can be stopped by: paper
    • So, the match is: Alpha (4/2 He) can be stopped by paper.

  2. 0/-1 e - This represents a beta particle.

    • Can be stopped by: aluminum
    • So, the match is: Beta (0/-1 e) can be stopped by aluminum.

  3. 0/0 y - This represents a gamma photon.

    • Can be stopped by: lead
    • So, the match is: Gamma (0/0 y) can be stopped by lead.

In summary, here are the matches:

  • Alpha (4/2 He) can be stopped by paper.
  • Beta (0/-1 e) can be stopped by aluminum.
  • Gamma (0/0 y) can be stopped by lead.
Match the solution to its description using the following diagram:

Column A
1.
20g of NaCl at 50°C:
20g of NaCl at 50°C
2.
45g KCl at 60°C:
45g KCl at 60°C
3.
70g of CaCl2 at 10°C:
70g of CaCl2 at 10°C
Column B
a.saturated
b.unsaturated
c.supersaturated

To match the solutions in Column A to their descriptions in Column B, we first need to understand the concepts of saturated, unsaturated, and supersaturated solutions:

  • Saturated: A solution that contains the maximum amount of solute that can be dissolved at a specific temperature. If more solute is added, it will not dissolve.

  • Unsaturated: A solution that can still dissolve more solute at a given temperature.

  • Supersaturated: A solution that has been prepared to contain more solute than can normally dissolve at that temperature; this often occurs when the solution is heated and then cooled.

Now, let's assign the descriptions based on typical solubility rules for the salts mentioned at their respective temperatures:

  1. 20g of NaCl at 50°C:
    Sodium chloride (NaCl) is quite soluble in water, and at 50°C, 20g of NaCl is generally considered to be in an unsaturated solution.
    Match: b. unsaturated

  2. 45g KCl at 60°C:
    Potassium chloride (KCl) is also highly soluble in water, and at 60°C, 45g is likely close to the saturation point. Assuming it exceeds 40g (which is commonly close to saturation), this would likely be saturated.
    Match: a. saturated

  3. 70g of CaCl2 at 10°C:
    Calcium chloride (CaCl2) is very soluble, and at 10°C, 70g would exceed typical solubility limits for water, making it likely a supersaturated solution.
    Match: c. supersaturated

Here are the final matches:

  • 1. 20g of NaCl at 50°C: b. unsaturated
  • 2. 45g KCl at 60°C: a. saturated
  • 3. 70g of CaCl2 at 10°C: c. supersaturated
In what group and period does Barium (Ba) belong?

a
Group 6, Period 2
b
Group 2, Period 6
c
Grroup 137, Period 56
d
Group 56, Period 137

Barium (Ba) belongs to:

b. Group 2, Period 6.

Here's the reasoning:

  • Group 2: Barium is an alkaline earth metal, which are located in Group 2 of the periodic table.
  • Period 6: Barium is also located in the sixth period of the periodic table.

So the correct answer is b. Group 2, Period 6.

For each description, decide if it is describing an acid or a base.

Column A
1.
Tastes sour.:
Tastes sour.
2.
Tastes bitter:
Tastes bitter
3.
H2SO4

:
H2SO4

4.
LiOH:
LiOH
5.
pH above 7:
pH above 7
6.
pH below 7:
pH below 7
Column B
a.Base
b.Acid

Here are the matches for each description from Column A with their corresponding classifications in Column B (acid or base):

  1. Tastes sour:

    • Description: This is typical of an acid.
    • Match: b. Acid

  2. Tastes bitter:

    • Description: This is typical of a base.
    • Match: a. Base

  3. H2SO4:

    • Description: Sulfuric acid (H2SO4) is a well-known strong acid.
    • Match: b. Acid

  4. LiOH:

    • Description: Lithium hydroxide (LiOH) is a strong base.
    • Match: a. Base

  5. pH above 7:

    • Description: A solution with a pH above 7 is considered a base.
    • Match: a. Base

  6. pH below 7:

    • Description: A solution with a pH below 7 is considered an acid.
    • Match: b. Acid

Here’s the summarized matching:

  • 1. Tastes sour: b. Acid
  • 2. Tastes bitter: a. Base
  • 3. H2SO4: b. Acid
  • 4. LiOH: a. Base
  • 5. pH above 7: a. Base
  • 6. pH below 7: b. Acid