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Anika [276]
3 years ago
6

In ionic bonds what happens to electrons

Chemistry
2 answers:
Nana76 [90]3 years ago
5 0

Answer:

C: metals donate electrons to non metals

Explanation:

bezimeni [28]3 years ago
3 0

Answer : The correct option is, (C) metals donate electrons to non metals

Explanation :

Covalent compound : Covalent compounds are the compound in which the atoms are covalently bonded.  The covalent bonds are formed by the equal sharing of the electrons and these bonds are formed between two non-metals.

Ionic compound : Ionic compounds are the compound in which atoms are bonded through the ionic bond.  The ionic bonds are formed by the complete transfer of electrons. That means the bond is formed when the metals donate electrons to the non-metals.

These bonds are formed between one metal and one non-metal.

Hence, the correct option is, (C) metals donate electrons to non metals

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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
Please help noowwwww... if you don’t know please don’t answer
pshichka [43]

Answer:

i honetly dont thing anyone knows that so look it up

Explanation:

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The reducing agent will itself be oxidized.
The oxidation number of carbon goes form 0 to +2. Therefore, it is the reducing agent.
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Answer:

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