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Degger [83]
3 years ago
11

Select the correct answer. The gas in a sealed container has an absolute pressure of 9.25 atmospheres. If the air around the con

tainer is at standard pressure, what is the gauge pressure inside the container? A. 0.759 atm B. 8.25 atm C. 10.25 atm D. 113 atm

Chemistry
2 answers:
N76 [4]3 years ago
7 0

Answer:

Explanation:

8.25 atm

yuradex [85]3 years ago
4 0

Answer: B) 8.25 atm.

Explanation: Absolute pressure is the sum of gauge pressure and atmospheric pressure.

P_a_b_s_o_l_u_t_e=P_g_a_u_g_e+P_a_t_m_o_s_p_h_e_r_e

Atmosphere pressure is given as 9.25 atm and the atmospheric pressure is 1.00 atm.

Let's plug in the values in the above formula:

9.25 atm = P_g_a_u_g_e + 1.00 atm

P_g_a_u_g_e = 9.25 atm - 1.00 atm

P_g_a_u_g_e = 8.25 atm

So, gauge pressure inside the container is 8,25 atm and hence the right option is B.

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How would the freezing point of seawater compare with that of pure water?
astraxan [27]
Freezing Point of Sea water would be lower than that of Pure Water. It is because of salinity of the water

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6 0
3 years ago
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A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c
nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0
3 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

8 0
3 years ago
(3 Points)
prisoha [69]
Your answer is B. radio waves have shorter wavelenghts than microwaves.
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5 0
3 years ago
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Which college classes would a student most likely take to become an oceanographer? Select 4 choices.
Setler [38]

Answer:

marine biology

coastal ecology

astronomy

meteorology

Explanation:

As a college student, to study oceanography one will have to take classes in the field of marine biology, coastal ecology, astronomy and meteorology.

An oceanographer is someone or a professional that studies the ocean in order have more scientific knowledge about them.

Oceanography is a merger of geology, biology, chemistry, physics as it pertains to the ocean.

  • There is no need to study human anatomy to study oceanography.
  • Marine biology and coastal ecology deals with study of life forms in their marine environment.
  • Astronomy and meteorology helps to gain insight about the formation of the ocean and how weather relates to the ocean.
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3 years ago
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