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3 years ago

What is the second step in the classification of living things?

Chemistry

2 answers:

frozen [14]3 years ago

8 0

Answer:

B.) Assigning them to any of the 3 domains

Explanation:

The second step in the classification of living beings is to assign them to any of the three domains. There are not six domains. Furthermore, there are neither six nor three kingdoms, there are 5 kingdoms.

The three domains are:

→ Bacteria Domain: It groups most of the unicellular prokaryotes, such as bacteria and cyanobacteria.

→ Domain Archaea: It groups prokaryotes with different biochemical characteristics and that usually inhabit regions with extreme conditions.

→ Eukaria domain: It groups all unicellular and multicellular eukaryotic organisms, such as fungi, animals, plants and organisms previously classified as protists.

Leya [2.2K]3 years ago

6 0

I'll go with B, have a good day

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Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur

Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of $C_5H_{12}$ = 21.9 g

Molar mass of $C_5H_{12}$ = 72.15 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_5H_{12}$ .

$\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles$

Now we have to calculate the moles of $H_2O$ .

The balanced chemical reaction will be,

$C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$

From the balanced reaction we conclude that

As, 1 mole of $C_5H_{12}$ react to give 6 moles of $H_2O$

So, 0.3035 moles of $C_5H_{12}$ react to give $0.3035\times 6=1.821$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$ .

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g$

Therefore, the mass of water produced will be 32.78 grams.

3 0

3 years ago

What can be said about an exothermic reaction with a negative entropy change?.

pashok25 [27]

Spontaneous at low temperatures.

3 0

2 years ago

If the density of an object is 5.2g/cm3, and it’s volume is 3.7 cm3, what is it’s mass?

lord [1]

Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3

5 0

3 years ago

A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

RoseWind [281]

Answer:

$C_3=0.125M$

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

$V_1C_1=V_2C_2$