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3 years ago

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In examining an organic compound, you discover that it contains the following functional group. What class of organic compound i

s it? O H || | R—C—N—R amide amine carboxylic acid ester

2 answers:

ki77a [65]3 years ago

6 0

the correct answer is a 100%

Harman [31]3 years ago

4 0

The answer is A. amide

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What is the predominant intermolecular force in the liquid state of each of these compounds: water (h2o), carbon tetrafluoride (

alexandr402 [8]

Water: hydrogen bonding
carbon tetrafluoride: dispersion or Van del Waals
dichloromethane: dipole-dipole

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3 years ago

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Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dis

Alexandra [31]

In order to find the molarity of the solution, we first require the moles of acetic acid added. For this,we need the mass which is:

Mass = volume * density

Mass = 50 * 1.05
Mass = 52.5 grams

Moles = mass / molecular weight

Moles = 52.5 / 60.05
Moles = 0.874 mol

Next, we know that the molarity of a solution is:

Molarity = moles / liter
Molarity = 0.874 / 0.5

Molarity = 1.75 M

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3 years ago

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What is meant by the term chemical reaction

pantera1 [17]

Answer:

<h3> $\huge{ \underline{ \underline { \bold{ \sf{ \orange{See \: below}}}}}}$ </h3>

Explanation:

▪️ $\underline{ \bold{ \sf{ \blue{Chemical \: reaction}}}}$

⇒The composition , decomposition or displacement of molecules of matter during chemical change is called chemical reaction.

▪️ $\underline{ \bold{ \sf{ \purple{ \: Further \: more \: explanation}}}}$

Various conditions bring about these changes. The chemical reactions are represented by chemicalequation. The compounds or elements that take part in chemical reaction are called reactant. They are written at the left side of an arrow that represent a change while the compound or elements that formed after the chemical change are called product. They are written at the right side of the arrow.

▪️ $\underline{ \bold{ \sf{ \purple{For \: example}}}}$

When nitrogen reacts with hydrogen to form ammonia :

Nitrogen + Hydrogen ⇒ Ammonia

N₂ + 3H₂ ⇒ 2NH₃

<u>Presentation </u><u>of </u><u>a </u><u>chemical </u><u>reaction </u><u>in </u><u>the </u><u>form </u><u>of </u><u>equation </u><u>is </u><u>called </u><u>chemical </u><u>equation </u>. <u>Chemical equation may be word equations or formula equations.</u>

Hope I helped!

Best regards!!

3 0

2 years ago

Which metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe? auau cucu mnmn agag

Cerrena [4.2K]

Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.

Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.

Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Learn more about electrode here:

brainly.com/question/17060277

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3 0

1 year ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

3 years ago