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skad [1K]
3 years ago
5

The geological time scale is divided up and organized according to what life forms existed on earth.

Chemistry
2 answers:
Ivanshal [37]3 years ago
8 0

Answer:

true

Explanation:

Geologic time is divided into four large segments called Eons: Hadean, Archean, Proterozoic, and Phanerozoic. The Phanerozoic Eon isdivided into Eras: Paleozoic, Mesozoic, and Cenozoic. The divisions among Eras reflect major changes in the fossil record, including the extinction and appearance of new life forms.

Pepsi [2]3 years ago
4 0

Answer:

TRUE

Explanation:

The geological time scale is divided on the basis of the information obtained from the<u> fossil evidences</u> and the rock specimens. The fossils of oldest organisms found on earth have depicted that the organisms were much simpler and with the increasing time, became more complex and structurally developed.

This classification was made by orderly arranging the main events that have taken place in the past, since the time of formation of the earth, till the present. The first unicellular life exited on earth about 3.5 billion years ago, and from that moment, continuous evolution has taken place depending upon the various geological factors that led to the development of more specific and well structured multi-cellular organisms by about 600-700 million years back.

Thus, the above statement is true.

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HACTEHA [7]
In comparison see it is very easy in goolge
4 0
3 years ago
If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
fredd [130]

Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0
3 years ago
Classify these definitions as that of an Arrhenius acid, an Arrhenius base, or other. Arrhenius acid definition Arrhenius base d
stealth61 [152]

Answer:

Explanation:

A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.

       This is an arrhenius Base

According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :

  • Sodium hydroxide NaOH
  • Potassium hydroxide KOH

A substance that produces an excess of hydrogen ion (H+) in aqueous solution

      This is an arrhenius Acid

An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.

Examples are;

  • Hydrochloric acid HCl
  • Hydroiodic acid HI
  • Hydrobromic acid HBr
5 0
3 years ago
a factor of 8, it means that A is now only 1/8 of its original concentration. A first order reaction, where A → products, has a
Trava [24]

Answer:

It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8

Explanation:

The equation that represents a first-order kinetics is:

Ln ([A] / [A]₀] = -kt

<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>

<em />

As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8

Replacing:

Ln ([A] / [A]₀] = -kt

Ln (1/8) = -1.57x10⁷s⁻¹*t

t = 1.32x10⁻⁷s

<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>

4 0
3 years ago
In chemical reactions, how does the valence configuration of bromine tend to change?
Strike441 [17]

Answer: Near the Protons. The electron structure of bromine is illustrated above. In chemical reactions, how does the valence configuration of Bromine tend to change? ... It loses one electron.

Explanation:

btw i found that on google lol

3 0
2 years ago
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