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Helga [31]
1 year 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]1 year 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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Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. ????????????H(????) → ????????+(?????
Dennis_Churaev [7]
Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)

Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
5 0
3 years ago
A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The ball
yanalaym [24]

Answer:

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Explanation:

Step 1: Given data

Initial volume of the balloon (V₁): 3.00 L

Initial pressure of the balloon (P₁): 765 torr

Final  volume of the balloon (V₂): ?

Final pressure of the balloon (P₂): 530 torr

Step 2: Calculate the final volume of the balloon

If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.

P_1 \times V_1 =  P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L

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