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3 years ago

Asolution has a pH of 8. Which best describes the solution?

2 answers:

7 0

A pH of 8 describes a weak base. Note that a pH of 7 is considered neutral. Anything lower than 7 is acidic, and anything higher than 7 is basic. For bases, the higher the pH, the more basic a substance is. Since the solution's pH of 8 is a lot closer to 7 than 14 (maximum pH), it is safe to assume that it is a weak base.

alekssr [168]3 years ago

6 0

Answer:

a weak base

Explanation:

pH levels:

0-3 Strong Acid

4-6 Weak Acid

7 Neutral

8-10 Weak Base

11-14 Strong Base

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The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th

andreyandreev [35.5K]

Answer:

The coefficient of thermal expansion α is

$\alpha = \frac{1}{T}$

The coefficient of compressibility

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

Explanation:

From the question we are told that

The coefficient of thermal expansion is $\alpha = \frac{1}{V} * (\frac{\delta V}{ \delta P}) P$

The coefficient of compressibility is $\beta = - (\frac{1}{V} ) * (\frac{\delta V}{ \delta P} ) T$

Generally the ideal gas is mathematically represented as

$PV = nRT$

=> $V = \frac{nRT}{P} --- (1)$

differentiating both side with respect to T at constant P

$\frac{\delta V}{\delta T } = \frac{ n R }{P}$

substituting the equation above into $\alpha$

$\alpha = \frac{1}{V} * ( \frac{ n R }{P}) P$

$\alpha = \frac{nR}{PV}$

Recall from ideal gas equation $T = \frac{PV}{nR}$

$\alpha = \frac{1}{T}$

Now differentiate equation (1) above with respect to P at constant T

$\frac{\delta V}{ \delta P} = -\frac{nRT}{P^2}$

substituting the above equation into equation of $\beta$

$\beta = - (\frac{1}{V} ) * (-\frac{nRT}{P^2} ) T$

$\beta =\frac{ (\frac{n RT}{PV} )}{P}$

Recall from ideal gas equation that

$\frac{PV}{nRT} = 1$

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

5 0

3 years ago

The first step in the formation of all three products is the loss of the ots leaving group to give a carbocation intermediate. t

lina2011 [118]

The carbocation stabilized by resonance structure and thereby lowers the energy of the carbocation, hydrogen will add to the carbon in the double bond that produces delocalization of electrons.

<h3>What is carbocation?</h3>

A carbocation is a molecule in which a carbon atom has a positive charge and three bonds.

In general, electrons are stabilized by delocalization. The stabilization energy engendered by delocalization over more than two atoms is called the resonance stabilization energy or simply the resonance energy. The greater the extent of electron delocalization the greater the resonance stabilization.

Learn more about the carbocation here:

brainly.com/question/19168427

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