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Kisachek [45]
3 years ago
8

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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Answer:

yes because the model showing you that it is the map

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2 years ago
The rate for a reaction between reactants l, m, and n is proportional to the cube of [l] and the square of [m]. what is the rate
postnew [5]
Answer
rate =k[l]^3[m]^2
4 0
3 years ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical rea
Aleksandr [31]

Answer:

ΔS° = - 47.2 J/mol.K

Explanation:

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ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10

∴ S°mH2O(l) = 69.9 J/mol.K

∴ S°mP4O10 = 231 J/mol.K

∴ S°mH3PO4 = 150.8 J/mol.K

⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231

⇒ ΔS° = - 47.2 J/mol.K

5 0
2 years ago
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Answer:

Nucleon

Explanation:

4 0
3 years ago
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