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Stels [109]
3 years ago
12

How does a buffer resist change in pH upon addition of a strong acid? The strong acid reacts with the strong base in the buffer

to form a salt, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH.
Chemistry
1 answer:
MrRa [10]3 years ago
4 0

Answer:

The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.

Explanation:

When a strong acid is added to the buffer, the acid dissociates and furnish hydrogen ions which combine with the conjugate of the weak acid, forming weak acid. The weak acid dissociates to only some extent and can furnish only some protons and there is no significant change in the pH.

Hence, option B is correct.

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The particles cannot move around at all. The particles are, however, still in motion.

3 0
3 years ago
A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?
Katen [24]

Answer:

T2 = 260 K  

Explanation:

<em>Given data:</em>

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K  

V1 = 1.75 L  

P2 = 210.0 kPa  

V2 = 1.30 L

<em>To find:</em>

T2 = ?

<em>Formula:</em>

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}

<em>Calculation:</em>

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K  

8 0
3 years ago
chegg write a net ionic equation describing the oxidation of no2 2 to no3 2 by o2 in a basic solution.
Marina86 [1]

When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.

NO2→NO^−3

NO2→NO

and do the usual changes

First, balance the two half reactions:

3. NO2 +H2O →NO^−3 + 2 H^+ + e−

4. NO2 +2 H^+ + 2e− → NO + H2O

Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:

5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−

Now add Eqn 4 and 5 (the electrons now cancel each other):

3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+

and cancel terms that’s common to both sides:

3NO2 + H2O → NO + 2NO^−3 + 2H+

This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.

Learn more about balancing equation here:

brainly.com/question/26227625

#SPJ4

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1 year ago
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If a stable element has an atomic number of 8, how many electrons does it have? HELP PLEASE!!!!!
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