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3 years ago

5

Draw a formula for Thr-Gly-Ala (T-G-A) in its predominant ionic form at pH 7.3. You may assume for the purposes of this question

that the pKa values of the acidic groups of amino acid residues in the peptide are the same as in the amino acid itself.

1 answer:

Rudiy273 years ago

8 0

Answer:

gggggggggg

Explanation:

gggggggg

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Is air a compound mixture or a element. compound

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Protons, neutrons, and electrons all have similar masses. explain why this statement is false.

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Which of the following is the electrical charge on an atom that contains 6 electrons, 3 neutrons, and 4 protons?

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3 years ago

Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so

dedylja [7]

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>

<em> </em>Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

<h3>3.18 (w/w) % </h3>

4 0

3 years ago

Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w

jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, $COCl_2$ , gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

The value of $K_c$ for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which $p_{CO}=p_{Cl_2}=0.265atm$ and $p_{COCl_2}=0.000atm$ ?

<u>Answer:</u> The equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of $K_c\text{ and }K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure

$K_c$ = Equilibrium constant in terms of concentration = 5.79

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side = $n_{g,p}-n_{g,r}=1-2=-1$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = Temperature = 570 K

Putting values in above equation, we get:

$K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124$

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

<u>Initial:</u> 0.265 0.265

<u>At eqllm:</u> 0.265-x 0.265-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}$

Putting values in above equation, we get:

$0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59$

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = $(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $Cl_2=(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $COCl_2=x=0.008atm$

Hence, the equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0

3 years ago