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eduard
2 years ago
6

Which organisms are extremely small and use hosts as both home and food supply?

Chemistry
2 answers:
Tasya [4]2 years ago
8 0
<span><u><em>Answer:</em></u>
Parasites

<u><em>Explanation:</em></u>
Parasites are a type of <u>pathogen</u> that survive by hitchhiking a host. They usually do this as they perceive the host as type of protection for them in order to survive.

Beside this protection, the host provides these parasites with the necessary nutrition as they feed directly from the host.
<u>They can feed on different parts from the hosts including:</u>
1- the digestion as in case of tapeworm
2- brain tissue as in case of brain-eating ameoba
3- blood

Parasites' presence in the host's body can alter the functionalities and processes within the body and cause sickness.

Hope this helps :)</span>
Simora [160]2 years ago
7 0
I would say that the bacteria who live inside one's gut would constitute very small organisms that are extremely small and inhabit the gut as home and also feed off it to survive and in this case probably have a symbiotic relationship with our bodies as they help us to break down and digest food.
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You have two compounds that you have spotted on the TLC plate. One compound is more polar than the other. You ran the TLC plate
goldenfox [79]

Answer:

we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

Explanation:

when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC  plate in a 50/50 mixture of hexanes and ethyl acetate

3 0
2 years ago
Please help!!
Phantasy [73]
1.
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7 0
2 years ago
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while your finger is still pushing the coin, there are four forces acting on the coin: what are they?
Viefleur [7K]
Is there some kind of diagram? how is your finger pushing the coin, and where? It may be:
1)friction against a surface
2)push from the finger
3)gravity
4)air resistance behind the coin
8 0
2 years ago
Read 2 more answers
Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F
nirvana33 [79]

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.  If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

  • <u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is False.

  • <u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)

Ionic form of the above equation follows:

Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

  • <u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is True.

  • <u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the correct answer is False.

  • <u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the correct answer is True.

3 0
2 years ago
You can identify a metal by carefully determining its density. A 8.44 g piece of an unknown metal is 1.25 cm long, 2.50 cm wide,
Dafna1 [17]

Answer: Aluminum, 2.70 g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.  It is characteristic of a substance.

Density=\frac{mass}{Volume}

Given : Mass of object = 8.44 grams

Volume of object=length\times breadth\times height=1.25cm\times 2.50cm\times 1.0cm=3.125cm^3

Putting in the values we get:

Density=\frac{8.44g}{3.125cm^3}=2.70g/cm^3

Thus density of the object will be 2.70 g/cm^3  which matches that of aluminium.

4 0
2 years ago
Read 2 more answers
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