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3 years ago

What is the correct formula of nickel (ii) oxide

2 answers:

7 0

Its simple: NiO
oxidation state of oxygen is -2
and that of nickel is given in the name itself as +2

so the compound would be simply NiO.

scoundrel [369]3 years ago

6 0

The correct formula for nickel (II) oxide is NiO.

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3. What is the substance ozone?

Reptile [31]

Chlorofluorocarbons and other halogenated ozone depleting substances are mainly responsible for man made chemical ozone depletion.

8 0

3 years ago

Why do electrons flow through a wire?

I am Lyosha [343]

Answer:

When electrons flow through a conductor such as a wire, it is called, "Electricity".

Explanation:

4 0

4 years ago

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

Finally, by removing the oxidation states we have:

$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

6 0

3 years ago

GuDViN [60]

Answer:

A phosphorus atom forms a P3- ion by gaining three electrons.

here

4 0

2 years ago

Determine the pH of a 5x10^-4 M solution of Ca(OH)2

miss Akunina [59]

Ca(OH)₂ ==> Ca²⁺ + 2 OH-

Ca(OH)₂ is strong Bases

Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻.

pOH = - log[ OH⁻]

pOH = - log [ 5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

pH = 10.7

hope this helps!

5 0

3 years ago