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vovangra [49]
3 years ago
10

If horizontal layers of sedimentary rock have a vertical fault running through them, how might a geologist use relative dating t

o determine when the fault occurred ? Could scientists use absolute dating as well ? Why or why not ?
Chemistry
2 answers:
Usimov [2.4K]3 years ago
7 0
1. If horizontal layers of sedimentary rock have a vertical fault running through them, then they cab predict where the fault begame by <span>matching up rock layers over distances.
2. Yes, because they it is use to </span><span>determine an age on a specified chronology in archaeology and geology. They can use these data to prove their theories.</span>
nata0808 [166]3 years ago
6 0

Answer:

Scientists will be able to use relative dating and absolute dating to determine when the failure occurred.

Explanation:

Scientists will be able to use relative dating, because it is the method capable of determining an order of events that occurred in the past by relating the rock with that failure to a rock without failure or with different failures, without the need to determine an absolute age of when the failure happened, however revealing a relative, approximate date.

It will also be possible to use absolute dating, which will give a more precise age on the fault present in the rock, using a chronological age that will be determined by observing the physical and chemical properties of the rock. This technique shows the age of the event in numbers, as opposed to relative dating.

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Read 2 more answers
For the following reaction, KcKc = 255 at 1000 KK.
bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

                           CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.      0.1550      0.173           0

At eqm.          (0.1550-x)  (0.173-x)         x

As we are given:

K_c=255

The expression for equilibrium constant is:

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get:

255=\frac{(x)}{(0.1550-x)\times (0.173-x)}

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M

The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

5 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
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