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3 years ago

Which type of rock might be useful as a source of petroleum?

Chemistry

2 answers:

fiasKO [112]3 years ago

8 0

Answer:

Geology of petroleum. Sedimentary rocks Petroleum may occur in any porous rock, but it is usually found in sedimentary rocks such as sandstone or limestone. Sedimentary rocks are grouped into three major classes: clastic, carbonate, and evaporitic.

Explanation:

storchak [24]3 years ago

6 0

Answer:

petroleum rocks

Explanation:

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Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the

Cloud [144]

Answer : The equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

Solution : Given,

Equilibrium constant, $K_c=1.8\times 10^{-5}$

Initial concentration of $HC_2H_3O_2$ = 0.260 m

Let, the 'x' mol/L of $H_3O^+$ are formed and at same time 'x' mol/L of $HC_2H_3O_2$ are also formed.

The equilibrium reaction is,

$HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)$

Initially 0.260 m 0 0

At equilibrium (0.260 - x) x x

The expression for equilibrium constant for a given reaction is,

$K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

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

$1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}$

By rearranging the terms, we get the value of 'x'.

$x=2.154\times 10^{-3}m$

Therefore, the equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

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3 years ago

It has been established that 8.0x10^4 tons of gold (Au) have been mined. Assume gold cost $350 per ounce. Whats the total worth

bazaltina [42]

So first lets find out how much gold there is. 10^4=10,000so that means 8.0x10,000= 8,000. now we have to find out how much money that is. All we have to do is multiply 8,000 by 350. When multiplied you get $2,800,000

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3 years ago

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Rudiy27

Answer:

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