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Murrr4er [49]
3 years ago
6

Conjugate acid/base problems: a. What are the conjugate bases of the molecules: i. C6H5OH? ii. CH3-SH iii. CH3-CH2-CO2H b. What

are the conjugate acids of the molecules: i. CH3–(CH2)-CO2- ii. CH3–(CH2)-NH2 iii. Ring at right 
Chemistry
1 answer:
DiKsa [7]3 years ago
6 0

Explanation:

In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species formed after the base accepts a proton. By contrast, a conjugate base is the species formed after an acid donates its proton.

Proton = H⁺

This means for the molecules that requires us to look for their conjugate bases, we simply remove a proton to it.

a. What are the conjugate bases of the molecules:

i C6H5OH : C6H5O⁻

ii. CH3-SH : CH3-S⁻

iii. CH3-CH2-CO2H : CH3-CH2-COO⁻

The molecules that requires us to look for their conjugate acids, we simply add a proton to it.

b. What are the conjugate acids of the molecules:

i. CH3–(CH2)-CO2- :  i. CH3–(CH2)-COOH

ii. CH3–(CH2)-NH2 : ii. CH3–(CH2)-NH3⁺

iii. Ring at right ?

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Atomic number is the number of protons present in an atom.

So, in isotopes the number of protons are same but the number of neutrons vary due to which atomic masses also vary.
In given three isotopes, all have same number of protons but different number of neutrons.
i.e.
H-1 = 1 P + 0 N = 1 u (Proton)
H-2 = 1 P + 1 N = 2 u (Deuterium)
H-3 = 1 P + 2 N = 3 u (Tritium)

Hence, it is clear that the number after H shows a change in number of neutrons and mass number.

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Hydrazine (N₂H₄), a rocket fuel, reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the e
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At STP one mol weighs 22.4L

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8 0
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32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac
Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0
3 years ago
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