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Stels [109]
3 years ago
7

Which kingdom does spiny lobster, and diatom belong to?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
3 0

Answer: The spiny lobster belongs to the kingdom ANIMALIA while diatoms belongs to the kingdom PROTISTA.

Explanation:

There are different ways biologists have used to examine and describe million kinds of living organisms. To classify organisms into groups, biologist look for important features that are shared by a large variety of organisms. These features which should have similar origin, structure and position. Organisms with certain important features in common are then put together into one group.

In the hierarchy of living organisms, the largest group of organisms is the KINGDOM. There are five types of kingdoms in which organisms are grouped, these includes:

--> kingdom Monera

--> kingdom PROTISTA,

--> kingdom fungi,

--> kingdom plantae and,

--> kingdom ANIMALIA.

Kingdom PROTISTA: they are single-celled, motile or non-motile organisms that are made up of eucaryotic cells, that is,cell structure is complex with a definite nucleus. Member include the DIATOMS, amoeba and chlamydomonas.

Kingdom ANIMALIA: They are multiple-celled, motile organisms that feed on other organisms. Their cells have no cell walls. The members include the SPINY LOBSTER, corals,worms, snails and cows

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True or false. for all atoms of the same element the 4s orbital is larger than the 3s orbital
Alenkinab [10]

Answer:

Between 4s and 3s orbital , 3s has more energy .

Explanation:

According to the rule , the lower the value of (n+l) for an orbital , the lower is it's energy . And if two orbitals have the same value of (n+l), the orbital with lower value of n will have the lower energy .

6 0
3 years ago
Can someone please help me
lilavasa [31]

Answer:

Z was smaller than X

Explanation:

if X<Y and Y<Z

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8 0
2 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Listed below are blood groups of o, a, b, and ab of randomly selected blood donors (based on data from the greater new york bloo
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The answer to this is 22, confirmed by gradpoint
3 0
3 years ago
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Consider a 5 molecule system. Assume that the molecules can either be found on the right side or left side of their container.
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Answer:

The correct answer is 6 possible states

Explanation:

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