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2 years ago

What is the IUPAC name for the compound shown below? ch3ch2ch2cnhch2ch3

Chemistry

2 answers:

antoniya [11.8K]2 years ago

6 0

Answer:

The name is Hexan-3-imine

Explanation:

The systematic name is formed by a prefix, which indicates the number of atoms of carbon that contains the molecule, and a suffix, which indicates the class of compound organic in question.

You must first look for the chain main, which is the chain with the highest amount of carbons. In this case the chain main contains six carbons. This means that the prefix used is Hex-

This case is an alkane (they have only single- bonds between carbons and hydrogen), so you use de suffix -ane

The carbon atoms of the main chain are numbered starting at the end closest to branching, so that the carbons with branches have the lowest possible number. In this case the branching is -NH, is in the position 3 and the name is imine.

So, knowing all this, you can name the compound like Hexan-3-imine

Nataly_w [17]2 years ago

3 0

We need the IUPAC name of the given compound.

The IUPAC name is: Hexan-3-imine.

The molecule has six carbon atoms in its skeleton. C=NH bond is attached to the skeleton at 3-position.

The functional group present in this molecule is imine (C=NH).

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How many molecules are there in 8.00 grams of glucose, C6H12O6

beks73 [17]

Answer:

molecules C6H12O6 = 2.674 E22 molecules.

Explanation:

from periodic table:

⇒ molecular mass C6H12O6 = ((6)(12.011)) + ((12)(1.008)) + ((6)(15.999))

⇒ Mw C6H12O6 = 180.156 g/mol

⇒ mol C6H12O6 = (8.00 g)(mol/180.156 g) = 0.0444 mol C6H12O6

∴ mol ≡ 6.022 E23 molecules

⇒ molecules C6H12O6 = (0.0444 mol)(6.022 E23 molecules/mol)

⇒ molecules C6H12O6 = 2.674 E22 molecules

6 0

3 years ago

A person accidentally swallows three drops of liquid oxygen, , which has a density of 1.149 g/ml. assuming the drop has a volume

DerKrebs [107]

134 ml First, let's determine how many moles of oxygen we have. Atomic weight oxygen = 15.999 Molar mass O2 = 2*15.999 = 31.998 g/mol We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml Since the density is 1.149 g/mol, we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2 Divide the number of grams by the molar mass to get the number of moles 0.17235 g / 31.998 g/mol = 0.005386274 mol Now we can use the ideal gas law. The equation PV = nRT where P = pressure (1.0 atm) V = volume n = number of moles (0.005386274 mol) R = ideal gas constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature ( 30 + 273.15 = 303.15 K) Now take the formula and solve for V, then substitute the known values and solve. PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L So the volume (rounded to 3 significant figures) will be 134 ml.

5 0

3 years ago

Read 2 more answers

The rate constant for this second‑order reaction is 0.190 M − 1 ⋅ s − 1 0.190 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product

Mamont248 [21]

Answer:

9.1 seconds

Explanation:

Given that for a second order reaction

1/[A]t = kt + 1/[A]o

Where [A]t= concentration at time = t= 0.340M

[A]o= initial concentration = 0.820M

k= rate constant for the reaction=0.190m-1s-1

t= time taken for the reaction (the unknown)

Hence;

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

5 0

2 years ago

Suppose you decide to define your own temperature scale using the freezing point (13 degrees C) and boiling point (360 degrees C

podryga [215]

M.P of oleic acid = 13 deg C
B.Pof oleic acid = 360 deg C 

(360 - 13) / 100 = 3.47 
So 1 deg O = 3.47 deg C 

The scale do not start at 0 deg C, it starts at 13 deg C. So to convert deg C to deg O,
subtract 13 then divide by 3.47 

deg O = (deg C - 13) / 3.47 

convert O to C multiply by 3.47 then add 13 

deg C = (deg O x 3.47) + 13 

convert 0 deg C to deg O 
deg O = (0 deg C - 13) / 3.47 
= - 3.75 deg O

8 0

3 years ago

What is the term for the breakdown of glucose in the absence of oxygen?

Tatiana [17]

Answer:

The breakdown of glucose in the absence of oxygen is termed as anaerobic respiration.

Explanation:

Anaerobic respiration or anaerobic oxidation occur in absence of oxygen molecule within the cytoplasm of cell that ultimately leads to the formation of lactate from pyruvate by the catalytic activity of lactate dehydrogenase enzyme.This reaction helps in the regeneration of NAD+.

During Anaerobic respiration only 2 molecules of ATP are formed by substrate level phosphorylation .

3 0

3 years ago