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nataly862011 [7]
3 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]3 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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the chloride of a metal M contains 47.25% of metal 1.0 gram of metal would be displaced from a compound by 0.88 gram of another
GenaCL600 [577]

Answer:

The equivalent weight of M is approximately 31.8 g

The equivalent weight of N is approximately 27.98 g

Explanation:

The given parameters are;

The percentage of the the metal M in in the chloride = 47.25%

Where by the chemical formula for the metal chloride is MClₓ, we have;

47.25% of the mass of MClₓ = Mass of M = W

Therefore, we have;

\dfrac{0.4725}{W} = \dfrac{1}{W + 35.5 \cdot x}

0.4725 × (W +  35.5·x) = W

0.4725·W + 0.4725×35.5×x = W

W - 0.4725·W  = 16.77·x

0.5275·W = 16.77·x

W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M

The equivalent weight of M = 31.799 ≈ 31.8 g

Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g

The equivalent weight of N = 27.98 g.

7 0
3 years ago
A welding tank of oxygen gas has a volume of 100. L. The oxygen in the tank is at a pressure of 95 atm and a temperature of 30.
Anuta_ua [19.1K]

Answer:

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5 0
4 years ago
Given 0.02 of KF, how many liters of solution are needed to make a 7.2 solution?
Dmitry_Shevchenko [17]

Answer:

1/360

Explanation:

let x = liters

molarity=moles of solute/liters of solution, 7.2=0.02/x or 7.2=(1/50)(1/x), 7.2(50)=(1/x), 360(x)=1, x=1/360

7 0
3 years ago
The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p
zysi [14]

Answer :    q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})  (in terms of mass)

or, q=\Delta H_{fusion}\times Moles     (in terms of moles)

Now we have to calculate the value of q.

q=6020J/mole\times 1Mole=6020J

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of \Delta e=0

Now we have to calculate the value of w.

Formula used :    \Delta e=q+w

where, q is heat required, w is work done and \Delta e is internal energy.

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

0=6020J+w

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 0
3 years ago
1. Consider the decomposition reaction of sodium chlorate. There are 100 grams of
IRINA_888 [86]
A. NaCl(s) and O2(g)

B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)

C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3

D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)

E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl

F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2

G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2

H. Percent yield = 10/45.1 • 100% = 22.2% yield
6 0
3 years ago
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