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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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A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a
Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

                 BaC₂O₄ ⇌   Ba²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

                ZnC₂O₄ ⇌   Zn²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0
3 years ago
What is the ground-state electron configuration of a neutral atom of manganese?.
Alchen [17]

Explanation:

elctronic configuration of manganese

Mn=1s²2s²2p⁶3s²3p⁶4s²3d⁵

ground state

Mn=Ar3d⁵4s²

note that Ar is argon

3 0
3 years ago
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