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Effectus [21]
3 years ago
10

The diagram below shows different layers of sedimentary rocks.

Chemistry
2 answers:
Flura [38]3 years ago
8 0
<span>Picture showing seven layers of rocks of different colors labeled A, B, C, D, E, F, and G from top to bottom;
A and B are parallel horizontal layers at the top of the diagram;
C, D, E, F, and G are slanted layers with C closest to the surface and G at the bottom.
</span>
The inference that is most likely correct is that (<span>C) Layer F is younger than Layer D.</span>
Artist 52 [7]3 years ago
3 0

Answer:

Layer F was formed earlier than Layer A.

Explanation:

i am taking the module exam on flvs and i am pretty sure its right because layer f is deeper than layer a so that means it is older or earlier than a if this is wrong please let me know and i apologize

btw i am tutoring in biology grades 6,7,8,9 for 5 dollars an hour so if you want to, just text me at 407-577-0617

i take cash app and paypal

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Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI
NeTakaya

Answer :

AgI should precipitate first.

The concentration of Ag^+ when CuI just begins to precipitate is, 6.64\times 10^{-7}M

Percent of Ag^+ remains is, 0.0076 %

Explanation :

K_{sp} for CuI is 1\times 10^{-12}

K_{sp} for AgI is 8.3\times 10^{-17}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

CuI\rightleftharpoons Cu^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][I^-]

1\times 10^{-12}=0.0079\times [I^-]

[I^-]=1.25\times 10^{-10}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgI\rightleftharpoons Ag^++I^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][I^-]

8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M

[Ag^+]=6.64\times 10^{-7}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{6.64\times 10^{-7}}{0.0087}\times 100

Percent of Ag^+ remains = 0.0076 %

8 0
3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yi
finlep [7]

Answer:

hypochlorite ion

Explanation:

The hypochlorous acid, HClO, is a weak acid with Ka = 1.36x10⁻³, when this acid is in solution with its conjugate base, ClO⁻ (From sodium hypochlorite, NaClO) a buffer is produced. When a strong acid as HCl is added, the reaction that occurs is:

HCl + ClO⁻ → HClO + Cl⁻.

Where more hypochlorous acid is produced.

That means, the HCl reacts with the hypochlorite ion present in solution

3 0
3 years ago
What is the most effective way to differentiate between potassium feldspar and plagioclase feldspar?
Tresset [83]
<span>plagioclase feldspars have striations and potassium feldspars don't have striations</span>
3 0
3 years ago
Which type of reaction is represented by the generic equation ab cd right arrow. ad cb? combustion decomposition single replacem
arsen [322]

The illustration would be that of a double replacement reaction.

<h3>What are double replacement reactions?</h3>

They are reactions in which 2 ionic compounds exchange ions to form two new products.

Thus, in the reaction: ab + cd ----------> ad + cb

ab and cd are two ionic compounds. The b in ab is replaced by the d in cd while the d in cd itself is replaced by the b in ab. Hence, new products, ad and cd, are formed.

More on double replacement reactions can be found here: brainly.com/question/19267538

#SPJ4

4 0
2 years ago
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