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1 year ago

When an aldose reacts with Barfoed's reagent, what type of organic compound forms? What type of chemical is this?

2 answers:

7 0

Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.

malfutka [58]1 year ago

3 0

Answer:

aldonic acid, Barfoed's reagent is copper(II) acetate in acetic acid.

Explanation:

Barfoed's test is the chemical test which is used for the detection of the presence of the monosaccharides.

The aldehyde group of aldose sugar is oxidized to the carboxylic group known as, aldonic acid.

RCHO + 2Cu²⁺ + 2H₂O → RCOOH + Cu₂O↓ + 4H⁺

Barfoed's reagent is copper(II) acetate in acetic acid which on reduces to copper(I) oxide forming a brick-red precipitate.

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An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine

nekit [7.7K]

Answer:

The molecular formula of the compound is $C_{7}H_{6}O_{2}$ .

Explanation:

Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:

Carbon

$m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{C} = 84.067\,g$

Hydrogen

$m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{H} = 6.057\,g$

Oxygen

$m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)$

$m_{O} = 31.995\,g$

Now, the number of moles ( $n$ ), measured in moles, of each element are calculated by the following expression:

$n = \frac{m}{M}$

Where:

$m$ - Mass of the element, measured in grams.

$M$ - Molar mass of the element, measured in grams per mol.

Carbon ( $m_{C} = 84.067\,g$ , $M_{C} = 12.011\,\frac{g}{mol}$ )

$n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }$

$n = 7$

Hydrogen ( $m_{H} = 6.057\,g$ , $M_{H} = 1.008\,\frac{g}{mol}$ )

$n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }$

$n = 6$

Oxygen ( $m_{O} = 31.995\,g$ , $M_{O} = 15.999\,\frac{g}{mol}$ )

$n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }$

$n = 2$

For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

$C_{7}H_{6}O_{2}$

8 0

1 year ago

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

1 year ago

Repulsion of electrons within two interacting molecules produces changes in electron distribution. This change in electron distr

ycow [4]

Answer:

Explanation:

This explains how two noble gases molecules can have an attractive force between them.

This force is called as van dar Waals forces.

It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .

If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles

6 0

1 year ago

What is the final acetate ion concentration when 69. g Ba(C2H3O2)2 is dissolved in enough water to make 970. mL of solution? The

Alexeev081 [22]

Answer:

0.56M

Explanation:

1. Dissociation equation

Assuming 100% dissociation, the equation is:

Ba(C₂H₃O₂)₂ → Ba²⁺ + 2C₂H₃O₂⁻

↑

acetate ions

2. Molarity

Calculate the molarity, M, of the solution:

M = n / V(in liters)

n = mass in grams / molar mass

n = 69.g / 255.415g/mol = 0.27015 mol

M = 0.27015mol / 0.970liter = 0.27850 mol/liter ≈ 0.28M

3. Acetate ions

From the chemical equation, 1 mol of dissolved Ba(C₂H₃O₂)₂ produces 2 acetate ions in solution.

Thus, 0.28 mol/liter × 2 = 0.56 mol/liter = 0.56M ← answer

8 0

1 year ago

At low temperatures and high pressures ethene gas C2H4(g), does not behave like an ideal gas. Use chemical principles to explain

Luba_88 [7]

Answer:

The particles of an ideal gas have no volume and no attractions for each other. In a real gas, however, the molecules do have a measurable volume. The molecules of real cases have intermolecular attractions for each other.

An ideal gas behaves like a real gas under the conditions of low temperature and high pressure.

This is because at low temperature and high pressure molecules of gas will have negligible kinetic energy and strong force of attraction.

Thus ethene gas does not behave like an ideal gas at low temperatures and high pressures.

6 0

1 year ago