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HACTEHA [7]
3 years ago
9

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 <em>pressure </em>of a gas increases, the <em>solubility </em>increases, and as the <em>pressure </em>of a gas decreases, the <em>solubility </em>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.

<u>Explanation:</u>

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.

8 0
3 years ago
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4 0
3 years ago
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