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2 years ago

QUestion 6 and 7

2 answers:

8 0

#6: The living organisms in an ecosystem can be divided into three categories: producers, consumers and decomposers. They are all important parts of an ecosystem.

#7: Nature has its own recycling system: a group of organisms called decomposers. Decomposers feed on dead things: dead plant materials such as leaf litter and wood, animal carcasses, and feces. They perform a valuable service as Earth's cleanup crew

HOPE THIS HELPS

Allisa [31]2 years ago

3 0

#7 decompsers eat on the dead. without decomposers dead leaves, dead animals and dead insects would pile up everywhere

You might be interested in

identify the reagents necessary to achieve each of the following transformations o3 dms, pcc, ch2cl, h2so4,h2o,hgso4

hoa [83]

the reagents necessary to convert alcohol to ketone $Na_2Cr_2O_7 , H_2O$ which involves oxidation of alcohols.

<h3>What is oxidation of alcohols?</h3>

Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
In order to cause primary alcohols to oxidize into aldehydes

$Cr_2O_7 ^-^2$ (dichromate)
$CrO_3$ /pyridine (Collins reagent)
Chromium pyridinium compound (PCC)
Dichromate of pyridinium (PDC, Cornforth reagent)
Periodinane by Dess-Martin
Oxalyl chloride with dimethylsulfoxide (DMSO) for Swern

oxidation of secondary alcohols to ketones

$Cr_2O_7 ^-^2$ (dichromate)
$CrO_3$ /pyridine (Collins reagent)
Chromium pyridinium compound (PCC)
Dichromate of pyridinium (PDC, Cornforth reagent)
Periodinane by Dess-Martin
Oxalyl chloride and dimethyl sulfoxide (DMSO) (Swern oxidation)
$CrO_3, H_2SO_4$ /acetone (Jones oxidation)
Acetone with aluminum isopropoxide (Oppenauer oxidation)

To learn more about oxidation of alcohols with the given link

brainly.com/question/7207863

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<u>Question:</u>

Identify the reagents necessary to achieve each of the following transformations

$a. O_3 , DMS$

$b.H_2SO_4 , H_2O , HgSO_4$

$c.Na_2Cr_2O_7 , H_2O$

$d.Fe ^ {2+}, NaOH$

3 0

2 years ago

Help me answer this please

Vlada [557]

<span>1=H, 2=B, 3=F, 4=A,5=C,6=E, 7=D, 8=G
</span>9: 69Ga=60.12% and 71Ga=39.88%; total=69.797amu
10: 27 27.977 92.23; 28 28.976 4.67; 29 29.974 3.10; abundance =28.07 Silicon

I hope this helps!

8 0

3 years ago

Why were chlorofluorocarbons first developed?

Sergio [31]

CFCs and their associated compounds were developed in the early 1900s as a non-toxic, non-flammable solution to other more dangerous products such as ammonia.

5 0

3 years ago

Direct electron transfer from a singlet reduced species to a triplet oxidizing species is quantum-mechanically forbidden

posledela

Direct electron transfer from a a singlet reduced species to a triplet oxidizing species is quantum-mechanically forbidden.

<h3><u>Transfer from singlet to triplet:</u></h3>

Either an excited singlet state or an excited triplet state will occur when an electron in a molecule with a singlet ground state is stimulated (through radiation absorption) to a higher energy level.
All electron spins in a molecule electronic state known as a singlet are coupled.
In other words, the ground state electron and the stimulated electron's spin are still coupled (a pair of electrons in the same energy level must have opposite spins, per the Pauli exclusion principle).
The excited electron and ground state electron are parallel in a triplet state because they are no longer coupled (same spin).
It is less likely that a triplet state would arise when the molecule absorbs radiation since excitation to a triplet state necessitates an additional "forbidden" spin transfer.

To view more questions on quantum mechanism, refer to:

brainly.com/question/13639384

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8 0

2 years ago

Consider the reaction:

goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0

3 years ago