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Trava [24]
3 years ago
8

Select all that apply. Which of the following bases are strong enough to deprotonate CH3CH2CH2Câ¡CH (pKa = 25), so that equilibr

ium favors the products?
a. CH3ONa
b. NaNH2
c. NaH
d. CH3CH2Li
e. H2O
f. NH3
Chemistry
1 answer:
Trava [24]3 years ago
5 0

Answer:

b, c, and d.

Explanation:

In an acid-base reaction, the equilibrium favors the weaker acid, thus, to the reaction favors it, the product must be weaker than the acid given. As higher is the value of pKa, as weak is the acid. Thus, the base must form a conjugate acid with pKa > 25.

The values of pKa can be found in equilibrium tables. The salts will dissociate to form the acids, which will gain one proton (H+)

a. CH3ONa -> Conjugate acid CH3OH, pKa = 15.5

b. NaNH2 -> Conjugate acid NH3, pKa = 38

c. NaH -> Conjugate acid H2, pKa = 35

d. CH3CH2Li -> Conjugate acid CH3CH3, pKa = 50

e. H2O -> Conjugate acid H3O+, pKa = -1.7

f. NH3 -> Conjugate acid NH4+, pKa = 9.2

Thus, the ones with pKa > 25 are NH3, H2, and CH3CH3, thus the bases are NaNH2, NaH, and CH3CH2Li.

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gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

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2 years ago
Analyze the graph below and answer the question that follows.
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C. (-2,-2) is the answer.

<u>Explanation:</u>

When we reflect a point (x, y) across the y-axis, after the reflection, the y-coordinate tends to be the same, however the x-coordinate is changed into its opposite sign.

Here U(2,-2) is reflected across the y-axis then,

the y-coordinate -2 remains the same and the x-coordinate is transformed into its opposite that is the sign of the x-coordinate will be changed as -2.

So the new coordinates of U after reflection will be (-2,-2).

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The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for
tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. K(s)+O_2(g)\rightarrow KO_2(s)

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g)

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D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g) : increases

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Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

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In reaction  N_2O_2(g)\rightarrow 2NO(g)+O_2(g) , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

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Silicon/Atomic number

14




8 0
2 years ago
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