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Answer:

#P_"O₂"# ≈ 100 mmHg in arterial blood, but it is different in other locations.

Explanation:

Here's a simplified diagram if the respiratory system.

Respiratory System
(From www.studyblue.com)

In ambient air, #P_"O₂"# = 160 mmHg.

In the alveoli

#P_"O₂"# in the alveoli is about 104 mmHg

The partial pressure in the alveoli is less than #P_"O₂"# in ambient air because of the continual diffusion of oxygen into the alveolar capillaries.

Leaving the alveolar capillaries

Oxygen diffuses from the alveoli into the alveolar capillaries. where #P_"O₂"# ≈ 100 mmHg.

In the pulmonary veins

There is no gas diffusion through veins and arteries, so #P_"O₂"# is about 100 mmHg.

Entering the systemic capillaries

Blood leaving pulmonary veins enters the left atrium and is pumped from the left ventricle into the systemic circulation.

It enters the systemic capillaries with #P_"O₂"# at 80 - 100 mmHg.

Leaving the systemic capillaries

#P_"O₂"# in the body cells is less than 40 mmHg.

Because #P_"O₂"# in the systemic capillaries is greater than the partial pressure in the body cells, oxygen diffuses from the blood and into the cells.

Leaving the systemic capillaries, #P_"O₂"# = 40 - 50 mmHg.

Entering the alveolar capillaries

Blood leaves the systemic capillaries and returns to the right atrium via veins.

The right ventricle then pumps the blood to the alveolar capillaries, with #P_"O₂"# = 20 - 40 mmHg, and the cycle starts again.

Here's an interesting animation showing the changes in #P_"O₂"# and #P_"CO₂"# as the blood moves through the body.

AHA! Actually, right after I wrote this question out, I figured it out.


The logic behind this is to express the change in pressure with respect to temperature at a constant volume in terms of #alpha# and #kappa#, since:

  • The container is rigid and full of liquid water. i.e. constant volume.
  • The water changes pressure due to the change in temperature. i.e. #(dP)/(dT) = ???#

Starting from the total derivative of the differential volume

#dV = ((delV)/(delT))_P dT + ((delV)/(delP))_T dP#,

we can divide by the partial differential temperature at a constant volume, #delT_V#, to find #((delP)/(delT))_V#.

#cancel(((delV)/(delT))_V)^(0) = ((delV)/(delT))_Pcancel(((delT)/(delT))_V)^(1) + ((delV)/(delP))_T stackrel("goal")overbrace(((delP)/(delT))_V)#

The term that cancels to #0# is because the change in volume must be #0# at a constant volume, and the term that cancels to #1# cancels because #(dT)/(dT) = 1#, and #1# at a constant volume is still #1#.

Thus, our big equation becomes:

#((delP)/(delT))_V = (-((delV)/(delT))_P)/((delV)/(delP))_T#

If we note that the definitions of #alpha# and #kappa# say #alpha = 1/V((delV)/(delT))_P# and #kappa = -1/V((delV)/(delP))_T#, then:

#((delP)/(delT))_V = (cancel(-V)alpha)/(cancel(-V)kappa) = alpha/kappa#

Now, we can do this trick where we multiply out a partial differential temperature at a constant volume to turn it into an exact differential, #dT_V#, and then integrate.

#dP_V = alpha/kappa dT_V#

#int_(P_1)^(P_2) dP_V = alpha/kappa int_(T_1)^(T_2) dT_V#

#P_2 - P_1 = alpha/kappa (T_2 - T_1)#

So the final expression is:

#bb(P_2 = P_1 + alpha/kappa (T_2 - T_1))#

And if we use this to evaluate the final pressure:

#color(blue)(P_2) = "1 atm" + (1.7xx10^(-4) "K"^(-1))/(4.7xx10^(-5) "atm"^(-1))("6 K")#

#= 22.7 ~~ color(blue)("23 atm")#

Answer:

WARNING! Long answer. (a) The C-terminus is glycine. (b) There are 4 charged groups at pH 7. (c) At pH 1 the net charge is +2. (d) The sequence using one-letter symbols is #"E-M-R-T-G"#.

Explanation:

(a) C-terminus

In a peptide, the amino acids are written from left to right with the #"NH"_2# group on the left and the #"C=O"# group on the right.

The left hand amino acid is called the N-terminus, and the right hand amino acid is called the C-terminus.

exploringtheworldofbiochem.files.wordpress.com

Thus, in #"Glu-Met-Arg-Thr-Gly"#, the C-terminus is Gly (glycine).

(b) Charged groups at pH 7

There are three points to remember:

  • If #"pH = p"K_a#, the neutral and ionic forms are present in equal amounts.
  • If #"pH < p"K_a# (i.e. more acidic), the protonated form will predominate.
  • If #"pH > p"K_a# (i.e. more basic), the non-protonated form will predominate.

Both Met (methionine) and Thr (threonine) have neutral R-groups, and they are neither N-terminal or C-terminal amino acids, so we can ignore them in our calculations.

Glu (glutamic acid) has both a basic N-terminus and an acidic R-group.

The corresponding #"p"K_"a"# values are

pH 7

The red line marks the pH 7 division mark.

For Gly, #"pH > pK"_"a"#, so the non-protonated form (#"COO"^"-"#) will predominate.

For Arg, #"pH < pK"_"a"#, so the protonated form (#"NH"_3^+#) will predominate.

For the R-group of Glu, #"pH > pK"_"a"#, so the non-protonated form (#"COO"^"-"#) will predominate.

However, for the N-terminal group of Glu, #"pH < pK"_"a"#, so the protonated form (#"NH"_3^+#) will predominate.

Thus, there are four ionized groups at pH 7.

(c) Net charge at pH 1

At pH 1, all groups have #"pH < pK"_"a"#, so the protonated forms will predominate..

pH 1

For the C-terminal group of Gly and the R-group of Glu, the protonated forms are #"COOH"#.

For the N-terminal group of Glu and the R-group of Arg, the protonated forms are #"NH"_3^+#.

Thus, at pH 1 the net charge is +2.

(d) Sequence using one-letter symbols

The one-letter symbols for the amino acids are

#"Arg = R"#
#"Glu = E"#
#"Gly = G"#
#"Met = M"#
#"Thr = T"#

#"Glu-Met-Arg-Thr-Gly = E-M-R-T-G"#.

A conceptual approach is to simply count electrons in a bond and treat each bonding valence electron as half a bond order.

This works for many cases, except for when the highest-energy electron is in an antibonding molecular orbital.

SIMPLE CASE

For example, the bond order of #:"N"-="N":# fairly straightforward because it's a triple bond, and each bonding valence electron contributes half a bond order.

So:

#"BO"_"triple" = "BO"_sigma + 2"BO"_pi = 1/2 xx ("2 electrons") + 2(1/2 xx ("2 electrons")) = 3# for the bond order, as we should expect, since bond order tells you the "degree" of bonding.

MULTI-ATOM CASE

Or, in a more complicated example, like #"NO"_3^(-)#, a conceptual approach is to count the number of electrons in the bond and see how many bonds it is distributed across.

http://i.stack.imgur.com/

So, in #"NO"_3^(-)#, which has one double bond in its resonance structure, has #2# electrons in its #pi# bond, distributed across three #"N"-"O"# bonds.

That means its #bb(pi)#-bond order is simply #1/2*("2 pi electrons")/("3 N"-"O bonds") = color(blue)("0.333")#, making the bond order for each #"N"-"O"# bond overall be:

#"BO" = "BO"_sigma + "BO"_pi = 1 + 0.333 = 1.333#.

Therefore, #"NO"_3^(-)# on average actually has three "#bb("1.333")#" bonds overall (instead of one double bond and two single bonds), meaning it is one third of the way between a single bond and a double bond.

EXCEPTION EXAMPLE: O2

#"O"_2# actually has two singly-occupied #pi^"*"# antibonding orbitals.

If we were to calculate its bond order, we would get #2# normally, corresponding to the #:stackrel(..)("O")=stackrel(..)"O":# Lewis structure.

But what if we wanted the bond order for #"O"_2^(+)#? From the discussion above, we may expect #1.5#, but it's NOT #1.5#, even though #"O"_2^(+)# has one less valence electron. What is it actually?

You may realize that we would have removed one electron from an #pi^"*"# antibonding molecular orbital. That means we've removed half a bond order corresponding to antibonding character, which is the same as adding half a bond order corresponding to bonding character.

So, by removing an antibonding electron, we've done the equivalent of adding a bonding electron.

In other words, we've decreased a bond-weakening factor, thereby increasing the bonding ability of the molecule.

Therefore, the actual bond order of #"O"_2^(+)# is #bb(2.5)#, stronger than #"O"_2#!

Answer:

Warning! Very long Answer! #"pH = (a) 9.26"; "(b) 5.23"; "(c) 4.51"; "(d) 3.13"#.

Explanation:

(a) pH at the beginning

We have an aqueous solution of the weak base, pyridine.

We set up an ICE table as usual.

#color(white)(mmmmmml)"C"_5"H"_5"N" + "H"_2"O" ⇌ "C"_5"H"_5"N"^"+" + "OH"^"-"#
#color(white)(mmmmmmmll)"Py" color(white)(ll)+ "H"_2"O" ⇌ color(white)(m)"PyH"^"+" color(white)(l)+ "OH"^"-"#
#"I/mol·L"^"-1":color(white)(mm)0.190color(white)(mmmmmmmll)0color(white)(mmmm)0#
#"C/mol·L"^"-1":color(white)(mml)"-"x"color(white)(mmmmmmml)+xcolor(white)(mml)+x#
#"E/mol·L"^"-1":color(white)(ml)"0.190-"x"color(white)(mmmmmmm)xcolor(white)(mmmm)x#

#K_b = (["PyH"^+]["OH"⁻])/"[Py]" = = (x·x)/(0.190-x) = x^2/(0.190-x) = 1.7 × 10^"-9"#

#0.190/(1.7 × 10^"-9") = 1.1 × 10^8 >400; ∴ x ≪ 0.190#.

#x^2/0.190 = 1.7 × 10^"-9"#

#x^2 = 0.190 × 1.7 × 10^"-9" = 3.23 × 10^"-10"#

#x = 1.80 × 10^"-5"#

#["OH"^"-"] = xcolor(white)(l) "mol/L" = 1.80 × 10^"-5"color(white)(l) "mol/L"#

#"pOH" = "-log"["OH"^"-"] = "-log"(1.80 × 10^"-5") = 4.74#

#"pH" = "14.00 - pH" = "14.00 - 4.74" = 9.26#

(b) After adding 12.5 mL of acid

#"Initial moles of Py" = 25.0 color(red)(cancel(color(black)("mL Py"))) × "0.190 mmolL Py"/(1 color(red)(cancel(color(black)("mL Py")))) = "4.750 mmol Py"#

#"Moles of HBr added" = 12.5 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "2.375 mmol"#

#color(white)(mmmmmm)"Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O"#
#"I/mol":color(white)(mm)4.750color(white)(ml)2.375color(white)(mmmml)0#
#"C/mol":color(white)(ml)"-2.375"color(white)(l)"+2.375"color(white)(mmll)+2.375#
#"E/mol":color(white)(mll)2.375color(white)(mml)0color(white)(mmmmml)2.375#

The solution contains 2.375 mmol of #"Py"# and 2.375 mmol of #"PyH"^"+"#.

We have a buffer.

#"pOH" = "p"K_"b" + log((["PyH"^"+"])/(["Py"])) = -log(1.7× 10^"-9") + log(stackrelcolor(blue)(1)(color(red)(cancel(color(black)"2.375 mmol")))/(color(red)(cancel(color(black)("2.375 mmol"))))) = 8.77 + 0 = 8.77.#

#"pH" = "14.00 - pOH" = "14.00 - 8.77" = 5.23#

(c) After adding 21 mL of acid

#" Moles of HBr added" =21 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "3.99 mmol H"_3"O"^"+"#

#color(white)(mmmmmm)"Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O"#
#"I/mol":color(white)(mm)4.750color(white)(mll)3.99color(white)(mmmml)0#
#"C/mol":color(white)(ml)"-3.99"color(white)(mm)"-3.99"color(white)(mmll)"+3.99"#
#"E/mol":color(white)(mll)0.76color(white)(mmm)0color(white)(mmmmll)3.99#

#"pOH" = "p"K_"b" + log((["PyH"^"+"])/(["Py"])) = -log(1.7× 10^"-9") + log((3.99 color(red)(cancel(color(black)("mmol"))))/(0.76 color(red)(cancel(color(black)("mmol"))))) = "8.77 + 0.72"=9.49.#

#"pH" = "14.00 -pOH" = "14.00 - 9.49" = 4.51#

(d) After adding 25 mL of acid

#" Moles of HBr added" =25 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) × ("0.190 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "4.75 mmol H"_3"O"^"+"#

#color(white)(mmmmmm)"Py" color(white)(ll)+ "H"_3"O"^"+" → color(white)(m)"PyH"^"+" color(white)(l)+ "H"_2"O"#
#"I/mol":color(white)(mm)4.750color(white)(mm)4.75color(white)(mmmml)0#
#"C/mol":color(white)(ml)"-4.75"color(white)(mm)"-4.75"color(white)(mmll)"+4.75"#
#"E/mol":color(white)(mmll)0color(white)(mmml)0color(white)(mmmmll)4.75#

We have a solution that contains 4.75 mmol of #"PyH"^"+"#.

#"Volume" = "25.0 mL + 25 mL" = "50 mL"#

#["PyH"^"+"] = "4.75 mmol"/"50 mL" = "0.095 mol/L"#

#color(white)(mmmmml)"PyH"^"+" + "H"_2"O" ⇌ color(white)(m)"Py"color(white)(l)+ "H"_3"O"^"+"#
#"I/mol":color(white)(mm)0.095color(white)(mmmmmmml)0color(white)(mmm)0#
#"C/mol":color(white)(mml)"-"xcolor(white)(mmmmmmmll)"+"xcolor(white)(mm)"+"x#
#"E/mol":color(white)(ml)"0.095-"xcolor(white)(mmmmmmll)xcolor(white)(mmm)x#

#K_a = (["Py"]["H"_3"O"^"+"])/(["PyH"^"+"]) = K_w/K_b = (1.00 × 10^"-14")/(1.7 × 10^"-9") = 5.88 × 10^"-6"#

#x^2/("0.095-"x) = 5.88 × 10^"-6"#

#0.095/(5.88 × 10^"-6") = 1.2 × 10^4 > 400; ∴ x ≪ 0.095#

#x^2/0.095 = 5.88 × 10^"-6"#

#x^2 = 0.095 × 5.88 × 10^"-6" = 5.59 × 10^"-7"#

#x = 7.48 × 10^"-4"#

#["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 7.48 × 10^"-4"color(white)(l) "mol/L"#

#"pH" = -log(7.48× 10^"-4") = 3.13#

Answer:

How about this?

Explanation:

Step 1.

The #δ^"-"# oxygen in a water molecule attacks the #δ^"+"# carbon in a #"CO"_2# molecule.

Step 1

Step 2.

A second water molecule deprotonates the newly-formed oxonium ion.

Step 2

Step 3

The newly formed hydronium ion transfers a proton to the negatively-charged oxygen atom that was formed in the first step.

Step 3

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