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PIT_PIT [208]
3 years ago
9

Provide one example of how life on Earth would not be possible without water based on its characteristics.

Chemistry
1 answer:
snow_lady [41]3 years ago
6 0

Answer: ice is less dense than liquid water. If ice was more dense, Earth would freeze.

Explanation: There are many reasons why life on Earth depends on the characteristics of water. One could discuss hydrogen bonds and its role as a solvent, but the unusual property of water is is the change in density with change in temperature. Water is densest at 4 degC, which is why ice floats - it is less dense than cold water (it melts quickly in warm water, so density isn’t impotant at higher temperatures). Most liquids are less dense than the solid, frozen form. If this was the case with water, any ice that formed would sink, and sease would freeze from the bottom up. Furthermore, the lowest layers would be insulated and would not all melt in summer. Thus over time, the seas would become a thin layer of liquid water at best, over solid ice. Life could not develop without liquid seas. In addition, ice is reflective, reducing the amount of sunlight absorbed, further reducing temperatures. Without ocean circulation, polar areas would be even colder, and there would be no rain.

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What mass of water was produced if 350.0 L of carbon dioxide were made at STP? C3H8 (g) + 5O2(g) → 2CO2 (g) + 4H2O (g)
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700 L of water was produced if 350.0 L of carbon dioxide were made at STP.

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4 0
2 years ago
A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and
Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

    The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = 99.7^oC

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

1atm=760mmHg

or,

1mmHg=\frac{1}{760}atm

As, 1mmHg=\frac{1}{760}atm

So, 752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

K=273+^oC

As, K=273+^oC

So, K=273+99.7=372.7

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor  = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}

\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole

Thus, the mass of one mole of vapor is 32.0 g/mole

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