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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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Which combination of elements are most likely to react to form ionic compounds
Nat2105 [25]
Ionic compounds form between metals and non metals.
7 0
3 years ago
What is the mass of 1.25 L of ammonia gas at STP
Alex73 [517]

Answer:

mass 1.25 Liters NH₃(gas) = 0.949 grams (3 sig-figs)

Explanation:

At STP (Standard Temperature-Pressure conditions => 0°C(=273K) and 1atm pressure,  1 mole <u>any</u> gas will occupy 22.4 Liters.

So, given 1.25 Liters ammonia gas at STP, convert to moles then multiply by formula wt. (17g/mole gives mass of NH₃.

moles NH₃(gas) = 1.25L NH₃(gas)/22.4L NH₄(gas)· NH₃(gas)mole⁻¹ = 0.0558 mole NH₃(gas).  

Converting to grams NH₃(gas) = 0.0558 mole NH₃(gas) x 17 g·mol⁻¹ = 0.949 grams NH₃(gas).

7 0
3 years ago
In the Haber process for the production of ammonia, what is the relationship between the rate of production of ammonia and the r
tangare [24]

Answer:

Δ[NH₃]/Δt  = 2/3 ( Δ[H₂]/Δt )

Explanation:

For determining rates as a function of reaction coefficients one should realize that these type problems are <u>always in pairs</u> of reaction components. For the reaction N₂ + 3H₂ => 2NH₃ one can compare ...

Δ[N₂]/Δt  ∝ Δ[H₂]/Δt, or

Δ[N₂]/Δt  ∝ Δ[NH₃]/Δt, or

Δ[H₂]/Δt  ∝ Δ[NH₃]/Δt, but never 3 at a time.

So, set up the relationship of interest ( ammonia rate vs. hydrogen rate)... nitrogen rate is ignored.

Δ[H₂]/Δt  ∝ Δ[NH₃]/Δt

Now, 'swap' coefficients of balanced equation and apply to terms given then set term multiples equal ...

N₂ + 3H₂ => 2NH₃   => 2(Δ[H₂]/Δt)  = 3(Δ[NH₃]/Δt) => 2/3(Δ[H₂]/Δt)  = (Δ[NH₃]/Δt)

NOTE => Comparing rates individually of the component rates in reaction process, the rate of H₂(g) consumption is 3/2 times <u>faster</u> than NH₃(g) production (larger coefficient).  So, in order to compose an equivalent mathematical relationship between the two, one must reduce the rate of the H₂(g) by 2/3 in order to equal the rate of NH₃(g) production. Now, given the rate of one of the components as a given, substitute and solve for the unknown.

CAUTION => When Interpreting rate of reaction one should note that the  rate expression for an individual reaction component defines 'instantaneous' rate or speed. <u>This means velocity (or, speed) does not have 'signage'</u>. Yes, one may say the rate is higher or lower as time changes but that change is an acceleration or deceleration for one instantaneous velocity to another. Acceleration and Deceleration do have signage but the positional instantaneous velocity (defined by a point in time) does not. Thus is reason for the  'e-choice' answer selection without the signage associated with the expression terms.  

3 0
2 years ago
Read 2 more answers
Define the terminology
Archy [21]
Biodiversity- variation of life on earth.. Such as economical status, genes, etc.
8 0
3 years ago
Suppose a 1.0 ml air bubble is trapped in the top of your buret and you do not notice it before measuring our liquid samples wit
Nadusha1986 [10]

The percentage error that would be introduced in a 35.0 mL sample of liquid is 2.86%

<h3>Data obatined from the question</h3>

From the question given, the following data were obtained:

  • Absolute error = 1 mL
  • True measurement = 35 mL
  • Perecentage error =?

<h3>How to determine the percentage error</h3>

The percentage error that would be introduced can be obtained as illustrated below:

Percentage error = (Absolute error / True measurement) × 100

Percentage error = (1 / 35) × 100

Percentage error = 2.86%

Thus, percentage error that would be introduced is 2.86%

Learn more about percentage error:

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7 0
1 year ago
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