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Minchanka [31]
2 years ago
14

g Synthesis of some disaccharides relies on the activation of glucose by UTP to form UDP‑glucose. First, UDP‑glucose is converte

d to UDP‑galactose. UDP - glucose UDP - galactose - 4 - epimerase ⇌ UDP - galactose Then, UDP‑galactose is combined with another glucose unit to form a disaccharide. Identify the disaccharide.
Chemistry
1 answer:
WITCHER [35]2 years ago
4 0

Answer: Lactose

Explanation:

Lactose is a sugar present in milk is produced from the glycosidic linkage of galactose and glucose. The anomeric carbon atom of beta-galactose is attached to the 4th hydroxy group of glucose through beta-1, 4 glycosidic linkage.

Galactose an epimer of glucose is produced in the initial steps when glucose is activated by UDP to form UDP-glucose it can be epimerized to Galactose.

When a galactose subunit is combined with another glucose units through a glycosidic linkage it forms a disaccharide Lactose.

You might be interested in
Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres
goldenfox [79]

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

3 0
2 years ago
Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben
sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O →  12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0
2 years ago
Calculate the pH of a polyprotic acid given and sketch the titration curves for the following reaction:
victus00 [196]

Answer: (1). pH = 1.70

(2). pH = 2.3

(3). pH = 3.3

(4). pH = 4.3

(5). pH = 8.41

(6). pH = 10.22

Explanation:

we assume that the formula representation of acid is H₂A

the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;

H₂A + OH⁻ → HA⁻ + H₂O

HA⁻ + OH⁻ → A²⁻ + H₂O

the volume of NaOH (V₀) at the first equivalent point is,

V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL

where volume of NaOH at 1/2 equivalent point is,

V₀/2 = 10.0mL

also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL

the volume of NaOH at 1/2 second equivalent point is,

V₀ + V₀/2 = 30.0mL

Volume of NaOH after second equivalence exceeds 40mL

therefore, at 0 mL NaOH addition;

where the extent of ionization is assumed to be x, we have

                        H₂A   ⇆     HA⁻   +   H⁺

where initial:   0.1 M       -            -

          change:   -x         +x           +x

          Equili:      0.1-x      x             x

Kаl = [HA⁻][H⁺] / [H₂A]

10⁻²³ = (x)(x) / (0.1-x)

x = 0.020

[H⁺] = 0.020 M

pH = -log [H⁺]

pH = -log(0.020)

pH = 1.70

(2). at 10 mL NaOH addition

[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol

[OH⁻] = 0.1 M * 10 mL = 1 mmol

after reaction:

[H₂A] = 1 mmol

[H⁻] = 1 mmol

pH = pKa₁ + log [HA⁻] / [[HA⁻]

pH = 2.3 + log 1mmol / 1mmol

pH = 2.3

(3). pH at the first equivalence point is,

pH = 1/2 (pKa₁ + pKa₂)

pH = 1/2(2.3 + 4.3) = 3.3

pH = 3.3

(4). pH at the second 1/2 equivalence point is

pH = pKa₂ = 4.3

pH = 4.3

(5). pH at the second equivalence point;

all H₂A is converted into A²⁻

[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M

at equilibrium:

                   A²⁻ + H²O    ⇆   HA⁺ OH⁻

          0.033 - x

from the Kb₁ expression,

Kb₁ = [OH⁻][HA⁻] / [A²]

Kw/Ka₂ = x²/(0.0333 - x)

10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)

x = 2.57 * 10⁻⁶

[OH⁻] = 2.57 * 10⁻⁶M

pH = -log Kw/[OH⁻] = 8.41

pH = 8.41

(6). pH after second equivalence point;

assuming the volume of NaOH is 40.10 mL

after second equivalence point OH⁻ in excess

[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M

pH = 0=-log Kw/[OH⁻] = 10.22

pH = 10.22

8 0
2 years ago
the mass of sample X is 20.0g. it was placed in a graduated cylinder and the water level rose from A to B. what is the density o
MatroZZZ [7]

Answer: 4.0 g/mL

Explanation:

The volume increased by 5.0 mL. Recall that the number of significant figures is equal to the number of certain values you can read plus one. Here, the volume increased from 14.0 mL to 19.0 mL, so the volume of X is 5.0 mL.

3 0
2 years ago
The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

The density of O₂ gas is 1.71 \frac{g}{L}

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

  • P=1.27 atm
  • molar mass of O₂= 32 \frac{g}{mol}.
  • R= 0.0821 \frac{atm*L}{mol*K}
  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
2 years ago
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