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3 years ago

What is the pOH of a 3.0 x10-3 M NaOH solution? pOH =

Chemistry

1 answer:

kodGreya [7K]3 years ago

4 0

Answer:

2.5

Explanation:

From the question given above, the following data were obtained:

Molarity of NaOH = 3.0x10¯³ M

pOH =?

Next, we shall determine the concentration of the hydroxide ion in the solution. This can be obtained as follow:

NaOH (aq) —> Na⁺ (aq) + OH¯ (aq)

From the balanced equation above,

1 mole of NaOH produced 1 mole of OH¯.

Therefore, 3.0x10¯³ M NaOH will also produce 3.0x10¯³ M OH¯.

Finally, we shall determine the pOH of the solution. This can be obtained as illustrated below:

Concentration of hydroxide ion [OH¯] = 3.0x10¯³ M

pOH =?

pOH = – Log [OH¯]

pOH = – Log 3.0x10¯³

pOH = 2.5

Thus, the pOH of the solution is 2.5

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A protein biochemist attempted to determine the amino acid sequence of a decapeptide. Use the results from the trypsin, chymotry

d1i1m1o1n [39]

Answer:

Explanation:

Before I start, let me make one thing clear before we start solving this...I know its says that it will not cleave if it's on the N-side of Proline. HOWEVER, this does not mean the WHOLE N-side...just the immediate N-side Amino Acid. This misunderstanding really screwed me over when I was trying to solve this question.

Ok, so let's start with 10 blanks:

X-X-X-X-X-X-X-X-X-X

Cyanogen bromide cleaves after methionine and leaves a nonapeptide so methionine must be the first one

Met-X-X-X-X-X-X-X-X-X

The information from the trypsin digestion lets us know the first part of the sequence. Since Trypsin cleaves the C side of Lys and Arg, Lys must be the last amino acid

Met-Val-Lys-X-X-X-X-X-X-X

CT3 has Lys, Met, Tyr, and Val and since Met, Val, and Lys are the first three amino acids, Tyr must be the fourth:

Met-Val-Lys-Tyr-X-X-X-X-X-X

Since chymotrypsin cleaves after Tyr and Trp and the sequence is only Thr-Trp, I just put it in next. This is kinda like a "gimme" sequence.

Met-Val-Lys-Tyr-Thr-Trp-X-X-X-X

(we don't have to worry for Trp to be next to Pro since we were told that Ala is the first Amino Acid of CT1's sequence)

All we are missing now is Ala, Arg, Phe, and Pro. Since Ala is the first amino acid of the CT1, that means:

Met-Val-Lys-Tyr-Thr-Trp-Ala-X-X-X

Now this is the part where it can be tricky to explain. This is spot where you can get multiple answers depending on where you look at first or what process of elimination you are using...

There are 4 possible ways to end this sequence...Phe-Arg-Pro, Arg-Phe-Pro, Arg-Pro-Phe, or Phe-Pro-Arg. HOWEVER, Arg-Phe-Pro and Phe-Arg-Pro CANNOT work.

Why?

For Phe-Arg-Pro:

If this was the case then there should have been a CT4: Arg-Pro

For Arg-Phe-Pro:

Then there should have been a T3: Phe-Pro

So your two answers can be:

Met-Val-Lys-Tyr-Thr-Trp-Ala-Phe-Pro-Arg OR Met-Val-Lys-Tyr-Thr-Trp-Ala-Arg-Pro-Phe

3 0

4 years ago

Give the outer electron configuration for each of the following columns in the periodic table. 1A,2A,5A,7A Express your answer a

Oliga [24]

Answer:

1A: ns^1

2A: ns^2

5A: ns^2np^3

7A: ns^2np^5

Explanation:

<u>According to IUPAC, the columns 1A, 2A, 5A, and 7A correspond to the groups, 1, 2, 15 and 17, respectively</u>.

The outer electron configuration of these columns are the following:

Column Outer electron configuration

1A ns^1

2A ns^2

5A ns^2np^3

7A ns^2np^5

Therefore, for the column 1A the s orbital has only one electron, for the column 2A has the s orbital completed with 2 electrons, for the column 5A the number of electrons in the s orbital is complete (2) and the number of electrons in the p orbital is 3, for the column 7A the s orbital has 2 electrons and the p orbital has 5 electrons.

I hope it helps you!

4 0

4 years ago

Calculate the pOH for a solution with a hydroxide ion, (OH-) of concentration of 6.49 X 10-11 M.

Ray Of Light [21]

Answer:

Calculate the pOH for a solution with a hydroxide ion, (OH-) of concentration of 6.49 X 10-11 M.

Calculate the pH.

Note: the answers should have three significant figures

The pOH is: Blank 1

The pH is: Blank 2

Explanation:

7 0

2 years ago

How many molecules (or formula units) are in each sample? 74.41 g ( CCl4 )

Alexus [3.1K]

Molar mass CCl₄ = 12 + 35.5 x 4 = 154 g/mol

number of moles:

74.41 / 154 => 0.48318 moles

1 mole ------------------- 6.02 x 10²³ molecules
0.48318 moles ----------- ?

molecules = 0.48318 x ( 6.02x10²³) / 1

molecules = 2.908 x 10²³ / 1

= 2.908 x 10²³ molecules

hope this helps!

3 0

3 years ago

Calculate the amount of heat needed to melt of solid octane () and bring it to a temperature of . Round your answer to significa

elixir [45]

Answer:

85.0 kJ

Explanation:

Calculate the amount of heat needed to melt 160. g of solid octane (C8H18 ) and bring it to a temperature of 99.2 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

To melt 160 g of octane and bring it's temperature to 99.2°C

(from literature)

Heat of fusion of Octane = 20.740 kJ/mol

Melting point of octane = -57°C

Boiling point of Octane = 125.6 °C

Molar mass of octane = 114.23 g/mol

Heat capacity of octane = 255.68 J/K.mol

So, it is evident that Octane is still a liquid at 99.2°C.

So, the required heat is the heat required to melt octane and raise the temperature of Octane liquid to 99.2°C

First, we convert the mass of octane given to number of moles as the heat parameters provided by literature are given in molar units.

Number of moles = (mass)/(Molar mass)

Number of moles of octane = (160/114.23) = 1.401 moles

Heat required to melt the octane = nL = (1.401×20.740) = 29.05674 kJ

Heat required to raise the temperature of already melted octane from its melting temperature of -57°C to 99.2°C

= nCΔT

n = 1.401 moles

C = 255.68 J/K.mol

ΔT = (99.2 - (-57)) = 156.2°C (same as a temperature difference of 156.2 K)

Heat required to raise the temperature of already melted octane from -57°C to 99.2°C

= (1.401×255.68×156.2)

= 55,952.04 J = 55.952 kJ

Total heat required to melt the 160 g of Octane and raise its temperature to 99.2°C

= 29.05674 + 55.952 = 85.01 kJ = 85.0 kJ

Hope this Helps!!!

8 0

3 years ago