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Helga [31]
10 months ago
9

What is the main difference between the Eubacteria and Archaebacteria kingdoms and the protista, fungi, plantae and Animalia kin

gdoms
Chemistry
1 answer:
gregori [183]10 months ago
5 0

The main difference between the Eubacteria and Archaebacteria kingdoms and the protista, fungi, plantae, and animal kingdoms is that Eubacteria and Archaebacteria are prokaryotes, while protista, fungi, plantae, and animal kingdoms are eukaryotes.

<h3>What is the Six Kingdoms Classification?</h3>

In the six kingdoms of classification, there are Eubacteria, Archaebacteria, Protista, Fungi, Plantae, and Animalia. The Eubacteria is a prokaryote, and its cell membrane is made up of peptidoglycan, teichoic acid, etc. Archaea are also prokaryotes, but their cell membrane composition differs from that of bacteria, so they are classified separately.

The eukaryotes are further classified into protista, fungi, plantae, and animalia depending upon their feeding patterns, complexity, etc. Protists may be unicellular or multicellular. Plants are autotrophs, fungi and animals are heterotrophs.

Eubacteria and Archaebacteria are examples of prokaryotes, whereas protists, fungi, plantae, and animals are examples of eukaryotes.

Learn more about the Six Kingdoms Classification, here

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List the number of each type of atom on the left side of the equation 2Na3PO4(aq)+2CoCl2(aq)→2Co3(PO4)2(s)+6NaCl(aq) Enter your
Trava [24]

Answer:

Na(sodium atom) = 6 atoms , P(phosphorus atom) = 2 atoms, Oxygen(Oxygen atom) = 8 atoms , Co(cobalt atom) = 2 atoms, Cl(chlorine atom) = 4 atoms

Explanation:

2Na3PO4(aq) + 2CoCl2(aq) → 2Co3(PO4)2(s) + 6NaCl(aq)

The question ask us to list the number of the kind of atom found on the left side(reactant) of the chemical equation. The question also reiterated that the order of the numbers should be the same as the order of the elements on the left side of the equation.

The type of atoms found on the left side of the equation are as follows;

Na(sodium atom) = 6 atoms

P(phosphorus atom) = 2 atoms

Oxygen(Oxygen atom) = 8 atoms

Co(cobalt atom) = 2 atoms

Cl(chlorine atom) = 4 atoms

The equation on the left hand side have sodium atom, phosphorus atom, oxygen atom, cobalt atom and chlorine atom.

Note: The given equation is not balanced but it can be balanced as follows;

2Na3PO4(aq) + 3CoCl2(aq) → Co3(PO4)2(s) + 6NaCl(aq)

The balanced equation have the following number of atom on the left side:

Na(sodium atom) = 6 atoms

P(phosphorus atom) = 2 atoms

Oxygen(Oxygen atom) = 8 atoms

Co(cobalt atom) = 3 atoms

Cl(chlorine atom) = 6 atoms

5 0
3 years ago
At room temperature, which of these are oxide minerals? Select all that apply.
DerKrebs [107]
C. and d. are oxyminerals.
7 0
3 years ago
Read 2 more answers
In your own words what is equilibrium?
fenix001 [56]

Answer:

when the forward and reverse reactions occur at equal rates.

chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.

8 0
3 years ago
What is the atmospheres in order from earth
lozanna [386]

Answer:

Earth's atmosphere is divided into five main layers: the exosphere, the thermosphere, the mesosphere, the stratosphere and the troposphere.

Explanation:

3 0
3 years ago
A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under
sukhopar [10]

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

8 0
2 years ago
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