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hichkok12 [17]
3 years ago
14

Which member of each pair is the stronger base? part a ethylamine or aniline ethylamine or aniline ethylamine aniline request an

swer part b ethylamine or ethoxide ion ethylamine or ethoxide ion ethylamine ethoxide ion?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0
Lowery-Bronsted theory is applied here. Acc. to this theory Base accepts protons and Acids donate proton.

Part 1: 
Aniline is less basic than ethylamine because the lone pair on nitrogen (which accepts proton) is not localized. It resonates throughout the conjugated system of phenyl ring. Hence due to unavailability of electrons for accepting proton it is less basic compare to ethylamine. In ethyl amine the lone pair of electron is localized and available to abstract proton.

Part 2:
In this case the alkyl groups attached to -NH₂ (in ethylamine) and -O⁻ (in ethoxide are same (i.e. CH₃-CH₂-). Ethoxide is more basic than ethylamine because ethoxide is a conjugate base of ethanol (pKa value of ethanol = 15.9 very weak acid) and the conjugate base of weak acid is always a strong base. Secondly, the oxygen atom more Electronegative than Nitrogen atom can attract more electron cloud from alkyl group as compared to Nitrogen in ethylamine. Hence, oxygen in ethoxide attains greater electron cloud than the nitrogen in ethylamine. Therefore, it is more basic than ethylamine.

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How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?
Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid \rm H_2SO_4 is a diprotic acid. When one mole of \rm H_2SO_4 molecules dissolve in water, two moles of \rm H^{+} ions would be produced.

\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}.

On the other hand, sodium hydroxide \rm NaOH is a monoprotic base. When one mole of \rm NaOH formula units dissolve in water, only one mole of hydroxide ions \rm OH^{-} would be produced.

\rm NaOH \to Na^{+} + OH^{-}.

Note that \rm H^{+} and \rm OH^{-} react at a one-to-one ratio:

\rm H^{+} + OH^{-} \to H_2O.

As a result, it would take 2\; \rm mol of \rm OH^{-} to react with the \rm 2\; mol of \rm H^{+} that was released when 1\; \rm mol of \rm H_2SO_4 is dissolved in water. Since one mole of \rm NaOH formula units could produce only one mole of \rm OH^{-}, it would take \rm 2\; mol of \rm NaOH formula units to produce that 2\; \rm mol of \rm OH^{-} for reacting with 1\; \rm mol of \rm H_2SO_4.

3 0
3 years ago
The three major types of faults are normal, reverse and syncline. true or false
Novosadov [1.4K]
The three major types of faults are Normal, Reverse and Strike-slip faults. 


Answer: FALSE
6 0
3 years ago
Read 2 more answers
HELP ASAP !!!!!
Anna35 [415]

Answer:

Explanation:

Take a random sample of nuts from the jar.  Let's take two handfuls, after shaking the jar and mixing the nuts thoroughly.  Separate the nuts into almonds and cashews.  Count each pile, then do the following calculation (these numbers are random, for example only).

                <u> Count</u>   <u>Percentage %</u>

Almonds       38        (38)/(87)x100

Cashews      <u> 49</u>         49/87x100

                      87          87/87 = 100%

Ratio of Almonds to Cashews:  <u>38/49</u>

3 0
2 years ago
When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem
masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

7 0
3 years ago
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)-&g
alexdok [17]

Answer:

<em>219 g</em>

Explanation:

6 0
3 years ago
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