1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nuetrik [128]
3 years ago
13

Enter an equation showing how this buffer neutralizes added aqueous acid (HI). Express your answer as a chemical equation. Ident

ify all of the phases in your answer.
Chemistry
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

H^++NH_3\rightleftharpoons NH_4^+

Explanation:

Hello there!

In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:

H^++NH_3\rightleftharpoons NH_4^+

Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.

Best regards!

You might be interested in
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
What is a limited resource in the use of nuclear fission for energy?
lys-0071 [83]
Uranium is the answer
5 0
3 years ago
Chemical and physical properties of calcite​
Semenov [28]

Answer:

Chemical Classification Carbonate

Explanation:

7 0
3 years ago
Jennie decides to mix the content of her beaker with another lab group to see “the results.” Is this safe why or why not explain
Ray Of Light [21]

Answer: no

Explanation:

It could be dangerous since you don't know what substance are you adding.it may end up in an explosion

I hope this helps :)

6 0
2 years ago
The figure below represents a reaction.What type of reaction is shown?
o-na [289]
Are you sure it isn’t SO3+H2O = H2SO4 because that would be combination (synthesis) A+ B=AB

Or SO3 + H2SO4 = H2S2O7
Because that would also be synthesis
7 0
3 years ago
Other questions:
  • Suppose a thin sheet of zinc containing 0.2 mol of the metal is completely converted in air to zinc oxide (zno) in one month. ho
    12·2 answers
  • A solution has a pH of 2.5. What is its [OH-] ?
    15·1 answer
  • What is the acceleration of the object’s motion? 0.5 m/s2 -0.5 m/s2 2 m/s2 -2 m/s2
    14·1 answer
  • What is produced each time a c-c bond is re-arranged or broken?
    13·1 answer
  • Graphite and iodine are non metal but they shine why secondry school
    14·1 answer
  • Beryllium exhibits diagonal relationship with- *<br> (a) Al<br> (b) B<br> (c) Mg<br> (d) Si
    7·1 answer
  • Chemical bonding may involve any of the following except the
    12·1 answer
  • Which of these types of changes is a physical change?
    11·2 answers
  • What makes an element the same as another element
    6·1 answer
  • Which would most likely result in low oxygen levels in the blood?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!