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kicyunya [14]
2 years ago
7

Viruses

Chemistry
2 answers:
sdas [7]2 years ago
8 0

Answer:

C are part of the Protista kingdom

Explanation:

The Protista, or Protoctista, are a kingdom of simple eukaryotic organisms, usually composed of a single cell or a colony of similar cells. Protists live in water, in moist terrestrial habitats, and as parasites and other symbionts in the bodies of multicellular eukaroytes.

Other eukaryotic kingdoms—the Plantae, Fungi, and Animalia—are each believed to be monophyletic . That is, all plants evolved from one ancestral plant, all animals from one ancestral animal, and all fungi from one ancestral fungus. The Protista, however, are not; they are almost certainly polyphyletic and did not arise from a single ancestral protist. Rather, the Protista are a category of miscellaneous eukaryotes, not closely related to each other and not sharing many characteristics, but not fitting any other kingdom of life. Some authorities divide the Protista into as many as twenty-seven phyla, and some feel the Protista should be discarded as a kingdom name, and these organisms divided into as many as twelve kingdoms.

Historically, the Protista were divided into three main categories: the plantlike algae, animal-like protozoans, and funguslike slime molds. This classification persists in many elementary textbooks; however, current molecular evidence indicates that these are not natural groups related by common descent, but groups with merely superficial , deceptive similarities.

Classifying them together is probably no more scientific than it would be to classify bees, birds, and bats in one group simply because they all have wings and fly. The two flagellated protozoan groups called trypanosomes and dinoflagellates, for example, are probably less related to each other than a human is to a fish. Genetic evidence (base sequences in their mitochondrial deoxyribonucleic acid [mtDNA] and ribosomal ribonucleic acid [rRNA]) now indicates that the following are more natural (evolutionarily related) groups of Protista.

These are the most primitive protists. Some lack mitochondria and suggest what the first eukaryotes may have been like, while others have primitive mitochondria that closely resemble bacteria. Some basal Protista without mitochondria are Trichomonas, a vaginal parasite of humans; Giardia, an intestinal parasite; and Entamoeba, the cause of amoebic dysentery. The lack of mitochondria is not necessarily the primitive (original) condition of all these protists, however. Although Giardia lacks mitochondria, it does have mitochondrial genes. Apparently it once had mitochondria, and these genes transferred to its nuclear DNA before the mitochondria were lost.

Basal Protista with mitochondria include Trypanosoma, a genus of blood parasites that cause African sleeping sickness and other diseases; Euglena, a green freshwater flagellated protozoan with chloroplasts; and Physarum, a common terrestrial slime mold.

pls give brainiest my hand hurts ;-:

Lelechka [254]2 years ago
7 0

Answer:

just see it it will help trust me its my school work

Explanation:

Download pdf
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What is the mass of CO2 lost at 20 min from the limestone sample?
harkovskaia [24]

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore  

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO​​​ 3

Hence in sample present CaCO​​​ 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO​​​3

8 0
3 years ago
The electronic configuration of an element is given below.
Shalnov [3]

Answer:

It is reactive because it has to gain an electron to have a full outermost energy level.

Explanation:

The electron configuration of oxygen is 1s2,2s2 2p4.

Oxygen is in group six in the periodic table so it has six electrons in its valence shell. This means that it needs to gain two electrons to obey the octet rule and have a full outer shell of electrons (eight).  

3 0
2 years ago
The azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib
stich3 [128]

Explanation:

Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.

The azide ion that is N^{-}_3, is a symmetrical ion, all of whose contributing structures have formal charges.

Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.

Contributing structures of azide ion are drawn in the image attached.

5 0
3 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ →  3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

  • Molar mass(MM) of Cr₂(SO₄)₃:

        = 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

        = 2*(52) + 3*(32) + 12*(16)

        = 104 + 96 + 192

        = 392 g

  • Molar mass(MM) of K₂SO₄:

        = 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

        = 2*(39) + 32 + 4*(16)

        = 78 + 32 + 64

        = 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : \frac{522 * 450}{392}

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0
3 years ago
Give 2 reasons why is copper used for electrial wires ​
adoni [48]

Answer:

Copper electrical wires are safer to use than wires made of most other conductive metals because they are resistant to heat. As you can see, copper is the preferred metal for electrical wires for several reasons. It has high electrical conductive; it's inexpensive; it's ductile; and it's thermal resistant.

4 0
3 years ago
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