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STALIN [3.7K]
4 years ago
10

Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a

solution containing 0.10 M NH4Cl and 1 M NH3. Buffer 2: a solution containing 1 M NH4Cl and 0.10 M NH3
Chemistry
1 answer:
Talja [164]4 years ago
5 0

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq).

\rm NH_3 gains one hydrogen ion to produce the ammonium ion \rm {NH_4}^{+}. In other words, \rm {NH_4}^{+} is the conjugate acid of the weak base \rm NH_3.

Both buffer 1 and 2 include

  • the weak base ammonia \rm NH_3, and
  • the conjugate acid of the weak base \rm {NH_4}^{+}.

The ammonia \rm NH_3 in the solution will react with hydrogen ions as they are added to the solution:

\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq).

There are more \rm NH_3 in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of \rm NH_3 in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

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Calculate the heat (in calories) required to freeze 35.0 g of water.
Serggg [28]

Answer:

-2.79 × 10³ cal

Explanation:

Step 1: Given data

  • Mass of water (m): 35.0 g
  • Latent heat of fusion of water (L): -79.7 cal/g

Step 2: Calculate the heat required to freeze 35.0 g of water

We have 35.0 g of liquid water and we want to freeze it, that is, to convert it in 35.0 g of ice (solid water), at 0 °C (melting point). We can calculate the heat (Q) that must be released using the following expression.

Q = L × m

Q = -79.7 cal/g × 35.0 g

Q = -2.79 × 10³ cal

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3 years ago
A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
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Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



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