Describe the evidence of changes in organisms in the Cambrian Era that are reflected in the fossil record.

Answers

Answer 1

Answer:

Answer:

Morphology and phylogenetics revealed by fossils. Perhaps the strongest evidence to support the Cambrian evolutionary explosion of animal forms is the first clear appearance, in the Early Cambrian, of skeletal fossils representing members of many marine bilaterian animal phyla

Explanation:

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Answer 2

Answer:

Final answer:

The Cambrian era saw significant changes in organisms reflected in the fossil record, including the emergence of complex multicellular life forms and the diversification of species. Fossils such as trilobites and brachiopods provide evidence of this biodiversity and adaptive radiation.

Explanation:

The Cambrian era is known for the rapid diversification of life forms, marked by the appearance of complex multicellular organisms. The fossil record from this era provides evidence of key evolutionary developments such as the emergence of hard-shelled animals, intricate body plans, and the colonization of different ecological niches. For example, the presence of trilobites, brachiopods, and archaeocyathids in the fossil record reflects the incredible biodiversity and adaptive radiation that occurred during the Cambrian.

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An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. This is the equation: CO(g)+H2O(g)--><-- CO2(g) + H2(g). How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answers

Answer : The moles of $CO_2$ added will be 1.12 mole.

Solution : Given,

Moles of $CO$ and $H_2O$ at equilibrium = 0.200 mol

Moles of $CO_2$ and $H_2$ at equilibrium = 0.600 mol

First we have to calculate the concentration of $CO,H_2O,CO_2\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CO=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }H_2O=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }CO_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

$\text{Concentration of }H_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$K_c=((0.600)* (0.600))/((0.200)* (0.200))$

$K_c=9$

Now we have to calculate the moles of $CO_2$ added.

Let the moles of $CO_2$ added is 'x'.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initially 0.200 0.200 0.600 0.600

Added moles 0 0 x 0

Change +0.1 +0.1 -0.1 -0.1

Final 0.3 0.3 (0.5+x) 0.5

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$9=((0.5)* (0.5+x))/((0.3)* (0.3))$

$x=1.12mol$

Therefore, the moles of $CO_2$ added will be 1.12 mole.

$\boxed{0.{\text{3 mol}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ are added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol.

Further Explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}} + {\text{Q(g)}} \rightleftharpoons {\text{R(g)}} + {\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}}=\frac{{\left[ {\text{R}}\right]\left[ {\text{S}}\right]}}{{\left[{\text{P}} \right]\left[ {\text{Q}} \right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

${\text{CO}}\left(g\right) + {{\text{H}}_2}{\text{O}}\left( g \right)\rightleftharpoons {\text{C}}{{\text{O}}_2}\left( g \right) + {{\text{H}}_2}\left( g \right)$

The expression for the equilibrium constant for the given reaction is as follows:

${\text{K = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}}\right]}}$ ......(1)

Here,

K is the equilibrium constant.

$\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ is the concentration of carbon dioxide.

$\left[{{{\text{H}}_{\text{2}}}} \right]$ is the concentration of hydrogen.

$\left[ {{\text{CO}}}\right]$ is the concentration of carbon monoxide.

$\left[ {{{\text{H}}_2}{\text{O}}}\right]$ is the concentration of water.

Substitute 0.600 mol/L for $\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ , 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}} \right]$ , 0.200 mol/L for $\left[ {{\text{CO}}}\right]$ and 0.200 mol/L for $\left[ {{{\text{H}}_2}{\text{O}}} \right]$ in equation (1).

$\begin{aligned}{\text{K }}&=\frac{{\left( {{\text{0}}{\text{.600 mol/L}}}\right)\left( {{\text{0}}{\text{.600 mol/L}}}\right)}}{{\left( {{\text{0}}{\text{.200 mol/L}}}\right)\left( {{\text{0}}{\text{.200 mol/L}}}\right)}}\n&= 9\n\end{aligned}$

The value of equilibrium constant comes out to be 9 and it remains the same for the given reaction.

Rearrange equation (1) to calculate .

$\left[{{\text{C}}{{\text{O}}_{\text{2}}}} \right]=\frac{{{\text{K}}\left( {\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}} \right]} \right)}}{{\left[ {{{\text{H}}_{\text{2}}}} \right]}}$ ......(2)

Substitute 9 for K, 0.300 mol/L for $\left[{{\text{CO}}}\right]$ , 0.200 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ and 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}}\right]$ in equation (2).

$\begin{aligned}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]&=\frac{{{\text{9}}\left( {{\text{0}}{\text{.300 mol/L}}} \right)\left( {{\text{0}}{\text{.200 mol/L}}} \right)}}{{{\text{0}}{\text{.600 mol/L}}}}\n&= 0.{\text{9 mol/L}}\n\end{aligned}$

Initially, 0.6 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ were present in a 1-L container. But now 0.9 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ are present in it. So the extra amount of ${\text{C}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of C}}{{\text{O}}_{\text{2}}}{\text{ added}} &= 0.{\text{9 mol}} - 0.{\text{6 mol}}\n&= 0.{\text{3 mol}}\n\end{aligned}$

Therefore 0.3 moles of carbon dioxide are added in a 1-L container.

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: CO, H2, CO2, H2O, 0.9 mol/L, 0.2 mol/L, 0.6 mol/L, 0.3 mol/L, K, carbon dioxide, water, hydrogen, carbon monoxide.

What is the volume of solution needed to make a 0.5M solution, using 250g of NaCl

Answers

1 mole of NaCl has (23+35.5)g= 58.5g

Therefore, to make 1M/L of solution you will need 58.5g Then to make 0.5M/L you will need 58.5/2=29.25g

If to make 1liter of 0.5M one will need 29.25 grams of NaCl, then250g of NaCl will make

(250g×1L)/29.25g= 8.547liters

The molar mass of NH3 is 17.03 g/mol. The molar mass of H2 is 2.0158 g/mol. In a particular reaction, 0.575 g of NH3 forms. What is the mass, in grams, of H2 that must have reacted, to the correct number of significant figures?

Answers

The balanced chemical reaction is:

N2 + 3H2 = 2NH3

We are given the amount of ammonia formed from the reaction. This is where we start our calculations.

0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2

Answer:

Answer is 0.102.

Explanation:

Which type of bond results when one or more valence electrons are transferred from one atom to another?(1) a hydrogen bond
(2) an ionic bond
(3) a nonpolar covalent bond
(4) a polar covalent bond

Answers

To solve this we must be knowing each and every concept related to an ionic bond. Therefore, the correct option is option 2 among all the given options.

What is an ionic bond?

Ionic bond, also known as electrovalent bond, is a kind of connection generated in a chemical molecule by the electrostatic attraction of oppositely charged ions.

Whenever the valence (outer portion) electrons with one atom are permanently transferred to another, a bond is formed.

The atom that loses electrons has become a positive ion (cation), whereas the atom that obtains electrons is becoming a negatively charged ion (cation) (anion). When one or more valence electrons are transferred from one atom to another, then the bond formed is an ionic bond.

Therefore, the correct option is option 2 among all the given options.

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The type of bond that results when one or more valence electrons are transferred from one atom to another is the ionic bond. The answer is number 2. The rest of the choices do not answer the question above.

17. Which of the following is true about an electrolytic cell? It changes electrical energy into chemical energy.
It is the type of cell used in electroplating
It uses an electric current to make a nonspontaneous reaction go.
All of the above
18. The products formed in the net reaction of the electrolysis of water are ____.
aqueous hydrogen ion and hydroxyl ion
hydrogen gas and oxygen gas
liquid hydrogen and oxygen gas
liquid oxygen and hydrogen gas
19. What occurs in electroplating?
decomposition of a metal layer
decomposition of a salt layer
deposition of a metal layer on a material
deposition of a salt layer on a metal

Answers

The correct answers are:

17. All of the above
18. hydrogen gas and oxygen gas
19. deposition of a metal layer on a material

Electrolytic cells use electric currents to drive a non-spontaneous reaction forward. This is also used in electroplating which involves the movement of metal cations to the surface of another metal. Water can also undergo electrochemical reactions by splitting it into its gaseous components, hydrogen and oxygen.

Final answer:

An electrolytic cell uses an electric current to make a nonspontaneous reaction go and is used in electroplating. The products formed in the electrolysis of water are hydrogen gas and oxygen gas. Electroplating involves the deposition of a metal layer on a material.

Explanation:

An electrolytic cell is a type of cell that uses an electric current to make a nonspontaneous reaction go. This means it can drive a chemical reaction in a direction that does not occur naturally. Electrolytic cells are commonly used in electroplating, where a metal layer is deposited onto a material. Therefore, all of the given options in question 17 are true about an electrolytic cell. For question 18, the products formed in the net reaction of the electrolysis of water are hydrogen gas and oxygen gas. Electrolysis of water involves the decomposition of water into its elemental components. In electroplating, the process involves the deposition of a metal layer on a material. This is achieved by passing an electric current through a solution containing metal ions, causing the metal ions to be reduced and deposit onto the material's surface.

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The density of an object is defined as mass per unit volume. Mathematically we write that: D = M/V. Rearrange this expression for: (a) M = ? (b) V = ?

Answers

The density of an object is defined as mass per unit volume. This can be rearranged to determine the mass or the volume of an object. For mass, the expression of density is rearranged to: M = VD. For the volume, the density is rearranged to: V = D/M.

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