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3 years ago

What is the boiling point of methanol, if ∆H°for vaporization is 35.3 kJ/mol, and ∆S° for vaporization is 105 J/mol K?

Chemistry

1 answer:

QveST [7]3 years ago

6 0

Answer: 336.2K

Explanation:

The boiling point is obtained from the standard enthalpy and entropy of vaporization as shown in the image attached.

Standard enthalpy of vaporization=35.3KJmol-1=35.3×10^3 Jmol-1

Standard entropy of vaporization= 105 JMol-1K-1

The formula and details of the solution are shown in the image attached.

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The solubility of a gas is 0.890 g/L at a pressure of 120 kPa. What is the solubility of the gas if the pressure is changed to 1

IgorC [24]

The solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

<h3>Effect of Pressure on Solubility </h3>

As the pressure of a gas increases, the solubility increases, and as the pressure of a gas decreases, the solubility decreases.

Thus, Solubility varies directly with Pressure

If S represents Solubility and P represents Pressure,

Then we can write that

S ∝ P

Introducing proportionality constant, k

S = kP

S/P = k

∴ We can write that

$\frac{S_{1} }{P_{1} } = \frac{S_{2} }{P_{2} }$

Where $S_{1}$ is the initial solubility

$P_{1}$ is the initial pressure

$S_{2}$ is the final solubility

$P_{2}$ is the final pressure

From the given information

$S_{1} = 0.890 \ g/L$

$P_{1} = 120 \ kPa$

$P_{2} = 100 \ kPa$

Putting the parameters into the formula, we get

$\frac{0.890}{120}=\frac{S_{2}}{100}$

$S_{2}= \frac{0.890 \times 100}{120}$

$S_{2}= 0.742 \ g/L$

Hence, the solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L

Learn more on Solubility here: brainly.com/question/4529762

7 0

2 years ago

Use the following table to answer the question. Strong acids Weak acids Weak bases Strong bases HBr CH3COOH NH3 NaOH HCl HF NH4O

Y_Kistochka [10]

Answer: The salt which has the highest pH is $CH_3COONa$

Explanation: Salts are formed by the reaction of acid and base. General equation for the generation of salt is given by:

$HX+BOH\rightarrow BX+H_2O$

The given salts are formed from the reaction of acids and bases.

1) $K_2SO_4$

This is a salt formed by the reaction of strong base (KOH) and strong acid $(H_2SO_4)$ . So, this salt is a neutral salt.

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

2) $CuSO_4$

This is a salt which is formed by the reaction of weak base $(Cu(OH)_2)$ and strong acid $(H_2SO_4)$ . So, this salt is an acidic salt.

$H_2SO_4+Cu(OH)_2\rightarrow CuSO_4+2H_2O$

3) $NH_4Cl$

This is a salt which is formed by the reaction of weak base $(NH_3)$ and strong acid (HCl). So, this salt is an acidic salt.

$HCl+NH_3\rightarrow NH_4Cl$

4) $CH_3COONa$

This is a salt which is formed by the reaction of strong base (NaOH) and weak acid $(CH_3COOH)$ . So, this salt is a basic salt.

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

Now, we have to find the salt which has the highest pH value.

Highest pH value means that the salt must be basic in nature and from the given salts, only $CH_3COONa$ is basic salt. Hence, this salt has the highest pH value.

6 0

3 years ago

Read 2 more answers

Please help me

Dmitriy789 [7]

Answer:

divergent

Explanation:

I believe it is divergent.

7 0

3 years ago

The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.47 D, where 1 D = 1 debye unit = 3.34 × 10-30 C-m. Ca

svetoff [14.1K]

The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 $\times$ 10^-5 V.

Explanation:

Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.

Given p = 1.47 D = 1.47 $\times$ 3.34 $\times$ 10^-30 = 4.90 $\times$ 10^-30.

V = 1 / (4π∈о) $\times$ (p cos(θ)) / (r^2)

where p is a permanent electric dipole,

∈ο is permittivity,

r is the radius from the axis of a dipole,

V is the electric potential.

V = 1 / (4 $\times$ 3.14 $\times$ 8.85 $\times$ 10^-12) $\times$ (4.90 $\times$ 10^-30 $\times$ 1) / (55.3 $\times$ 10^-9)^2

V = 1.44 $\times$ 10^-5 V.

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3 years ago