Hg + O2 --> HgO reaction

fish tank initially contains 35 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 5 l

weeeeeb [17]

Answer:

Therefore, the rate of change in the amount of salt is $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$

$\frac{grams }{min}$

Explanation:

Given:

Initial volume of water $V = 35$ lit

Flowing rate = 5 $\frac{Lit}{min}$

The rate of change in the amount of salt is given by,

$\frac{dx}{dt} =$ ( Rate of salt enters tank - rate of sat leaves tank )

Since tank is initially filled with water so we write that,

$x(0) = 0$

Let amount of salt in the solution is $c$ ,

$\frac{dx}{dt} = \frac{5c}{1 } - \frac{x(t) \times 5}{100}$

$\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$ $\frac{grams}{min}$

Therefore, the rate of change in the amount of salt is $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$

$\frac{grams }{min}$

7 0

3 years ago

Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met,

erma4kov [3.2K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

Explanation:

The acid-catalyzed hydrolysis from makes us to understand that the polypeptides contains 8 amino acids

From the question we are told that the Edman's reagent releases Leu it means that the N-terminal amino acid would be Lue(i.e Leucine)

Also from the question we are told that the Carboxypeptidase released Ser this mean the the C-terminal amino acid would be Ser ( i.e Serine)

The Amino acids would in the polypeptides would be arranged like this

Lue __ __ __ __ __ __ Ser

Now from the question we are told that treatment with cranogen bromide form two peptides.

Now generally cranogen bromide divides a peptide on the C- side(i.e the extreme left side ) of Met(Methionine)(This is an amino acid obtained by hydrolysis of most common peptides )

So this means that any peptides containing Met(Methionine) must have Methionine as a C- terminal amino acid(i.e at extreme left) and for peptides that does not contain Met must be C - terminal peptides

From the question we see that it is the second peptide that contain Met and it is a penta peptide(i.e it contains 5 amino acid)

Thus the fifth amino acid is Met

So the sequence of these amino acid would now be

Lue __ __ __ Met __ __ Ser

From the question we are told that the the Trypsin-catalyzed hydrolysis forms two amino acid and two peptides

Now generally Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine) and Lys (lysine) and any peptide that holds Arg or Lys must have them as their C- terminal amino acids

From the first peptide in the two peptide formed we see that Arg would be the Seventh amino acid of the octapeptide because commonly the trypsin that sticks to the C-side of Arg would for Ser

and Phe would be the sixth amino acid of the octapeptide

So the sequence of these amino acid would be

Lue __ __ __ Met Phe Arg Ser

Looking at the first amino acid formed from the Trypsin-catalyzed hydrolysis we see that Arg would be the fourth amino acid of octapeptide as Trypsin divides a peptide on the C- side(i.e the extreme left) of Arg(Arginine).

From the second peptide of the Trypsin-catalyzed hydrolysis we see that Lys would be the third amino acid of the octapeptide as trypsin divides on the C- side of Lys (lysine) and Tyr would be the second amino acid of the octapeptide

So the sequence of these amino acid would be

Lue Arg Lys Arg Met Phe Arg Ser

4 0

4 years ago

How many grams of ethane gas (C2H6) are in a 12.7 liter sample at 1.6 atmospheres and 24°C? Show all work used to solve this pro

dexar [7]

n = PV/RT

p = 1.6 atm

v = 12.7L

R = 0.0821

T = 24°C which is equivalent to 297.15 degrees k

n = (16 × 12.7) / (0.0821 × 297.15)

n = 20.32 / 24.39

n = 0.83 mol

C = 12.90

H = 1.0079

C2 = 12.010 × 2 = 24.02

H6 = 1.0079 × 6 = 6.0474

C2H6 = 30.0674

Ethane times n which is 30.0674 × 0.83mol

= 24.95 grams of C2H6. Which is Ethane.

4 0

4 years ago

What is 5 ounces in milliliters?

SIZIF [17.4K]

1 ounces ----------- 29.5735 mL
5 ounces ----------- ??

5 x 29.5735 / 1 => 147.868 mL

hope this helps!

6 0

3 years ago

Please answer this question it’s due in 5 minutes thanks if you do ! :))

Oksi-84 [34.3K]

Answer:

you are correct

Explanation:

3 0

3 years ago