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olga nikolaevna [1]
2 years ago
10

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4

will be needed to produce 725 mL of a solution that has a concentration of Na ions of 1.30 M
Chemistry
1 answer:
balu736 [363]2 years ago
8 0

Answer:

51.53 grams .

Explanation:

Na₃PO₄    ⇄  3Na⁺¹   +    PO₄⁻³ .

1 mole           3 mole

725 mL of 1.3 M Na⁺  ions

= .725 x 1.3 moles of Na⁺ ions

= .9425 moles

3 mole of Na⁺  is formed by 1 mole of Na₃PO₄

.9425 mole of Na⁺  is formed by .9425/3  mole of Na₃PO₄

Na₃PO₄ needed =  .9425/3  moles = .3142 moles

Molecular weight of Na₃PO₄ = 164

grams of Na₃PO₄ needed = .3142 x 164 = 51.53 grams .

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Answer:

  • <em>As the temperature of a sample of matter is increased, the average kinetic energy of the particles in the sample </em><u>increase</u><em>.</em>

Explanation:

The <em>temperature</em> of a substance is the measure of the <em>average kinetic energy </em>of its partilces.

The temperature, i.e. how hot or cold is a substance, is the result of the collisions of the particles (atoms or molecules) of matter.

The kinetic theory of gases states that, if the temperature is the same, the average kinetic energy of any gas is the same, regardless the gas and other conditions.

This equation expresses it:

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Where Avg KE is the average kinetic energy, R is the universal constant of gases, N is Avogadro's constnat, and T is the temperature measure in absolute scale (Kelvin).

As you see, in that equation Avg KE is propotional to T, which means that as the temperature is increased, the average kinetic energy increases.

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Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
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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

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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 )

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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

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