What is the second hottest layer of the atmosphere?

How does a buffer resist change in ph upon addition of a strong acid?

lora16 [44]

Any buffer exists in this equilibrium

HA <=> $H^+ + A^-$
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base ( $A^-$ )

When a strong acid is added, it reacts with the large reservoir of the conjugate base ( $A^-$ ) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.

8 0

3 years ago

Valance shell example..

Natali5045456 [20]

Answer:

<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2>

<h3>Valence electrons are the electrons in the outermost shell, or energy level, of an atom. </h3>

<h3>Configuration of oxygen's valence electrons as 2s²2p⁴.</h3>

Explanation:

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3 0

2 years ago

how to determine the net charge of the tripeptide Asp-Gly-Leu at pH 7. Can someone show in details and tricks on how to solve it

Ugo [173]

Answer:

0!

Explanation:

You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:

-COOH -NH3

pH 7------------------------------------------------------

Asn - +

Gly - +

Leu - +

Now that we have our table we'll sketch our peptide's structure:

This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:

+1 (HN-Asn)

-1 (Leu-COOH)

=0

The net charge is 0!

I hope you find this information useful and interesting! Good luck!

5 0

3 years ago

What energy is using the telephone?

kakasveta [241]

Answer:

electrical

Explanation:

i kinda guessed.

4 0

3 years ago

Read 2 more answers

Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no rea

iragen [17]

Answer :

(A) The balanced molecular equation will be:

$K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)$

(B) The balanced molecular equation will be:

$Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)$

(C) The balanced molecular equation will be:

$Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)$

(D) The balanced molecular equation will be:

$Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}$

Explanation :

Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.

Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

Part A : potassium carbonate and lead(II) nitrate

The balanced molecular equation will be:

$K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)$

In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.

Part B : lithium sulfate and lead(II) acetate

The balanced molecular equation will be:

$Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)$

In this reaction, lead sulfate is an insoluble salt and lithium acetate is a soluble solution.

Part C : copper(II) nitrate and sodium sulfide

The balanced molecular equation will be:

$Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)$

In this reaction, Cuprous sulfide is an insoluble salt and sodium nitrate is a soluble solution.

Part D : strontium nitrate and potassium iodide

The balanced molecular equation will be:

$Sr(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+SrI_2(aq)$

In this reaction, strontium iodide and potassium nitrate are soluble solution.

$Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}$

6 0

3 years ago