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Gekata [30.6K]
3 years ago
11

A gravel can be broken down into different kinds of substances by physical processes. No chemical reactions are needed to separa

te different parts of gravel into pure substances. This is because gravel is a(n)
Chemistry
2 answers:
maxonik [38]3 years ago
8 0

Answer:

D) heterogeneous mixture

Explanation:

the other person just put mixture but did not specify what type, so for those who need this, here you go :)

Vlada [557]3 years ago
6 0

Answer:

MIXTURE

Explanation:

A mixture is a substance composed of a combination of other different substances. These component(s) of a mixture are physically combined, meaning that there is no chemical linkage between the individual components/constituents of a mixture.

This is the case of the gravel described in this question. The components of gravel can be separated using physical means because they are not chemically bonded to one another, hence, no chemical reactions are needed to separate different parts of gravel into pure substances. This makes gravel a MIXTURE.

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Water molecules are polar, with the Select one:
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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
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