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2 years ago

11

2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.

Tyrosine

1 answer:

Savatey [412]2 years ago

5 0

<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

Glutamic -----> 3.08

Asparagine -----> 5.43

Tyrosine -----> 5.63

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

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if element X has two valence electrons and element Y has five valence electrons the formula for the compound is?

malfutka [58]

Answer:

the answer to the question is X3Y2

7 0

2 years ago

Suppose a 500.mL flask is filled with 1.0mol of CO, 1.5mol of H2O and 0.70mol of CO2. The following reaction becomes possible: +

valentina_108 [34]

Answer:

[CO] = 0.62 M

Explanation:

Step 1: Data given

Volume of the flask = 500 mL

Number of moles CO = 1.0 moles

Number of moles H2O = 1.5 moles

Number of moles CO2 = 0.70 moles

The equilibrium constant K for this reaction is 3.80

Step 2: The balanced equation

CO(g) + H2O(g) ⇆ CO2(g) +H2(g)

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CO] = 1.0 moles / 0.500 L = 2.0 M

[H2O] = 1.5 moles / 0.500 L = 3.0 M

[CO2] = 0.70 moles / 0.500 L = 1.4 M

[H2] = 0M

Step 4: The concentration at the equilibrium

For 1 mol CO we have 1 mol H2O to produce 1 mol CO2 and 1 mol H2

[CO] = 2.0 -X M

[H2O] = 3.0 - X M

[CO2] =1.4 + X M

[H2] = X M

Step 5: Define Kc

Kc = [CO2][H2]/ [CO][H2O]

3.80 = (1.4 + X) * X / ((2.0 - X)*3.0 -X))

X = 1.38

[CO] = 2.0 -1.38 = 0.62 M

[H2O] = 3.0 - 1.38 = 1.62 M

[CO2] =1.4 + 1.38 M = 2.78

[H2] = 1.38 M

Kc = (2.78*1.38) / (0.62*1.62)

Kc = 3.8

[CO] = 0.62 M

8 0

3 years ago

How many mL of 0.100 M NaCl would be required to make a 0.0365 M solution of NaCl when diluted to 150.0 mL with water?

Tatiana [17]

Answer:

54.75 mL

Explanation:

First calculate the number of moles of NaCl in the 150mL solution of NaCl

0.0365 moles should be present on 1000cm3 or 1dm3 of water.

1L = 1 dm3

1 mL = 1 / 1000 dm3

150 mL = 150/1000 dm3 = 0.15 dm3

If x moles are present in 0.15 dm3,

x/ 0.15 = 0.0365

We get x= 0.0365 × 0.15 mol

Now x amount of moles should be taken from the initial 0.100 M NaCl solution

So 0.1 moldm-3 = 0.0365× 0.15 mol / V

we get V = 0.05475 dm3

V= 0.05475 L

V= 54.75 mL

3 0

3 years ago

1. Convert the following:<br> 5 meters =<br> kilometers

a_sh-v [17]

It equals 0.005 but plz mark this as the brainliest question plz

5 0

4 years ago

Read 2 more answers

Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic ac

Elza [17]

Answer:

$H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1$

$H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2$

$HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3$

Explanation:

Hello!

In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:

$H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1$

$H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2$

$HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3$

Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:

Ka1 > Ka2 > Ka3

Best regards!

3 0

3 years ago