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nataly862011 [7]
2 years ago
13

The vapor pressure of Substance X is measured at several temperatures:temperature (C) vapor pressure (atm)34 0.23644 0.29254 0.3

55Use this information to calculate the enthalpy of vaporization of X. Round your answer to 2 significant digits. Be sure your answer contains a correct unit symbol. Clears your work. Undoes your last action. Provides information about entering answers.
Chemistry
1 answer:
shusha [124]2 years ago
6 0

Answer:

ΔHv = 17.04 KJ/mol

Explanation:

T(°C)   T(K)      Pv(atm)         1/T(K)                LnPv

34      307      0.236       0.00325733      - 1.4439

44      317       0.292       0.0031545         - 1.2310

54      327      0.355       0.003058          - 1.0356

Clausius-Clapeyron:

  • δLnP/δT = ΔH/RT²

⇒ δLnP = ΔH/R (δT/T²)

∴ δT/T² = δ/δT(- 1/T )

⇒ δLnP/δT = - ΔH/R

Graphing: LnP vs 1/T

we get an ecuation that corresponds to a straight line:

y = - 2049.6x + 5.2331 ...... R² = 1

where the slope of this line is:

y = mx + b

⇒ m = - 2049.6 = - ΔH/R.....Clausius-Clapeyron

⇒ ΔH = (2049.6)(R)

∴ R = 8.314 E-3 KJ/mol.K

⇒ ΔHv = 17.04 KJ/mol

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Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode (m_{ch}) is given as :

\frac{m_{ch}}{M} ZF = It

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

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also; (m_{ch}) = (Ad) \rho ; replacing that into above equation; we have:

\frac{(Ad) \rho}{M} ZF = It  ---- equation (1)

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From the above equation (1); The time required for coating can be calculated as;

[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons }  ] = (20 A) t

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8 0
2 years ago
A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight
zhannawk [14.2K]

Answer: The molecular formula will be C_8H_8O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{5.9}{1.5}=4

For H = \frac{5.9}{1.5}=4

For O =\frac{1.5}{1.5}=1

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is C_4H_4O

The empirical weight of C_4H_4O = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2

The molecular formula will be=2\times C_4H_4O=C_8H_8O_2

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Answer:

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