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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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How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
What was the robotic vehicle Opportunity designed to do?
mestny [16]
I believe the answer is C
6 0
4 years ago
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What is the density of an object that has a mass of 20g and a volume of 10 cc?
Alexxandr [17]
i believe
2 g/mL
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5 0
3 years ago
Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?
faltersainse [42]

Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

3 0
3 years ago
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