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3 years ago

Solid mercury (II) oxide produces liquid mercury and oxygen gas

1 answer:

7 0

Mercury(II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas. Mercury(II) oxide is a red solid. When it is heated, it decomposes into mercury metal and oxygen gas. A reaction is also considered to be a decomposition reaction even when one or more of the products are still compounds.

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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

When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol

oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, $O_2$

Solution : Given,

Moles of $FeCl_3$ = 3.00 mole

Moles of $O_2$ = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)$

From the balanced reaction we conclude that

As, 2 moles of $FeCl_3$ react with 1 mole of $O_2$

So, 3.00 moles of $FeCl_3$ react with $\frac{3.00}{2}=1.5$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $FeCl_3$ is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, $O_2$

4 0

2 years ago

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Alenkinab [10]

Answer:

Explanation:

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3 years ago

What is a small description of a atom

steposvetlana [31]

Answer:

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3 years ago

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Which is the electron configuration for lithium

creativ13 [48]

Answer:

The electron configuration is [HE] 2s1

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3 years ago