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puteri [66]
3 years ago
8

When the surface water temperature is still well below the water boiling temperature, some bubbles at the bottom tend to float u

p but somehow the bubbles seem to disappear before reaching the surface. What is the most likely scenario for these bubbles?
Chemistry
1 answer:
irga5000 [103]3 years ago
7 0

Answer:

The reasons why the seemingly floating bubbles disappear was that they tend to loss their latent heat to the water molecules at the surface water.

Explanation:

Heat energy has a considerable effect on the velocity of molecules including water. The water molecules below the container will receive much more heat energy than those above it. This heat energy in the form of specific heat capacity and latent heat that result in the increase in the speed of individual molecules of water and finally to the escape of the molecules to a colder region of the container, in this case the upper region. At the collision of the bottom water to the surface water, they tend to exchange their heat content, the hotter molecules will lose their heat to the cold ones. When the formerly hot molecules encounter this, it will result in lowering the temperature and consequentially to the reduction of their movement, once in the form of bubble, now become ordinary water. This convectional transfer of heat energy will continue until the whole system has a uniform temperature depending on the consistency of the heat source.

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

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At what Celsius temperature does 0.750 mol of an ideal gas occupy a volume of 35.9 L at 114 kPa?
shepuryov [24]

382.85  Celsius is the temperature does 0.750 moles of an ideal gas occupy a volume of 35.9 L at 114 kPa.

Explanation:

Given data:

number of moles of the gas  = 0.75 moles

volume of the gas = 35.9 liters

pressure of the gas = 114 KPa or 1.125 atm

R = 0.0821 latm/moleK

temperature of the gas T = ?

The equation used to calculate temperature from above data is ideal gas law equation.

the equation is :

PV = nRT

T = \frac{PV}{nR}

Putting the values in the above rewritten equation:

T = \frac{1.125 X 35.9}{0.75 X 0.0821}

T = 655.9 K

To convert kelvin into celsius, formula used is

K = 273.15+ C

putting the values in the equation

C = 656 - 273.15

   = 382.85  Celsius

8 0
3 years ago
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