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jek_recluse [69]
3 years ago
14

You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second jar

there is a small amount of water. explain how the vapor pressure of water in the first jar compares with the vapor pressure of water in the second jar.
Chemistry
1 answer:
melisa1 [442]3 years ago
3 0

since both the jars are kept at the same temperature the vapor pressure will be same in both the cases.


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A sample of gas occupies 9.0 mL at a pressure of 500.0 mm Hg. A new volume of the same sample is at a pressure of 750.0 mm Hg.
Anuta_ua [19.1K]
<h3>Answer:</h3>

                The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

<h3>Solution:</h3>

              According to Boyle's Law, " <em>The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure</em>". Mathematically, the initial and final states of gas are given as,

                                     P₁ V₁  =  P₂ V₂    ----------- (1)

Data Given;

                  P₁  =  500 mmHg

                  V₁  =  9.0 mL

                  P₂  =  750 mmHg

                  V₂  =  ??

Solving equation 1 for V₂,

                   V₂  =  P₁ V₁ / P₂

Putting values,

                   V₂  =  (500 mmHg × 9.0 mL) ÷ 750 mmHg

                   V₂  =  6.0 mL

<h3>Result:</h3>

            The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

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1 year ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

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3 years ago
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