Calculate the molality of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4) (35.4 % means 35.4 g of H3PO4in 100 g of solution)

Answer: The molal boiling point elevation constant of X is

Explanation:

Formula used for Elevation in boiling point :

or,

where,

= boiling point constant  = ?

m = molality

= mass of solute (urea) = 55.4 g

= mass of solvent  X =  500 g

= molar mass of solute (urea) = 60 g/mol

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

Thus the molal boiling point elevation constant of X is

Answer:

CaCl2 + H2O  Ca2+(aq) + 2Cl-(aq)

Explanation:

CaCl2 + H2O  Ca2+(aq) + 2Cl-(aq)

When CaCl2 is dissolved in H2O (water) it will dissociate (dissolve) into Ca+2 and Cl- ions.

The dissolution of calcium chloride is an exothermic process.  

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

Final answer:

The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.

Explanation:

The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.

The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.

With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.

For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.

Learn more about Thermodynamics of Styrene Production here:

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The total pressure of the gas, in kPa is equal to 102.4 kPa.

Given the following data:

• Atmospheric pressure = 100.8 kPa

• Pressure (open side) = 12 mm Hg

To determine the total pressure of the gas, in kPa:

First of all, we would convert the value of the pressure in mm Hg to kPa.

Conversion:

760 mm Hg = 101.325 kPa

12 mm Hg = X kPa

Cross-multiplying, we have:

X = 1.60 kPa

Now, we can calculate the total pressure by using this formula:

Total pressure = 102.4 kPa

Read more: brainly.com/question/22480179

Answer:

102kPa

Explanation:

760mmHg = 101.325 kPa

12mmHg = x

Upon converting from mmHg to kPa we have;

x = ( 12 mm Hg x  101.325 kPa)  / 760mmHg

x = 1.599868421 kPa

Total pressure = 100.8 kPa + 1.599868421 kPa

Total pressure = 102kPa

Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃  +  3H₂  →   3H₂O  +  W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten(VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten(VI) oxide would need (0.194  .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass  . mol)

0.194 mol . 183.84 g/mol = 35.6 g