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laiz [17]
3 years ago
6

Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).

Chemistry
1 answer:
likoan [24]3 years ago
5 0

Answer:

1.40*10⁻² M

Explanation:

We have the solubility formula

Solubility,

S = KH*P  

where

KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm

P = atmospheric pressure = 0.400 atm

Hence, we have

S = KH*P

= (3.50*10⁻² mol/L.atm)*(0.400 atm)

= 1.40*10⁻² mol/L

But 1 mol/L = 1 M,

Hence, the answer (1.40*10⁻² mol/L ) is equivalent to

= 1.40*10⁻² M

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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. C2H6 + O2 → CO2 + H2O
fenix001 [56]
Step 1: Write the unbalanced equation,

                                 C₂H₆  +  O₂    →    CO₂  +  H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

                                 C₂H₆  +  O₂    →   2 CO₂  +  H₂O

Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.


                                 C₂H₆  +  O₂    →   2 CO₂  +  3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

                                 C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

                           ( C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O ) × 2

                                2 C₂H₆  +  3 O₂    →   4 CO₂  +  6 H₂O
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Explanation:

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This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable

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