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3 years ago

What are the two major types of cells that occur in nature?

Chemistry

2 answers:

cupoosta [38]3 years ago

5 0

Answer:

Cells are of two types: eukaryotic, which contain a nucleus, and prokaryotic, which do not. Prokaryotes are single-celled organisms, while eukaryotes can be either single-celled or multicellular.

Explanation:

asambeis [7]3 years ago

3 0

if you get it wrong i'm sorry i'm just a dumb person

Cells are of two types: eukaryotic, which contain a nucleus, and prokaryotic, which do not. Prokaryotes are single-celled organisms, while eukaryotes can be either single-celled or multicellular.

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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

Hi, im 15 i live in indiana and i need friends please i love tacos and im an all around fun person but very emotionally drained.

Andrei [34K]

Answer:

you sound like a cool guy, ill email ya one of these days

Explanation:

4 0

2 years ago

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When a strontium atom loses two electrons to form an Sr2+ ion, the electrons are lost from the

kherson [118]

Answer:When a strontium atom loses two electrons, it becomes a(n) cation with a charge of 2+. when any neutral atom loses an electron it becomes cation that is positively charged ions and an ion gets the charge according to the number of electrons loses.

you are very welcome but I may be inaccurate.

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3 years ago

Use the correct terms to explain the osmotic concentrations of the solutions.

Morgarella [4.7K]

The movement of water from lower concentration to higher concentration trough semipermeable membrane.

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3 years ago

1. All non-living components of an ecosystem form a(n)

zubka84 [21]

Abiotic environment

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2 years ago

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