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sashaice [31]
3 years ago
9

Balance the reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coefficient is "1.

" H₂O₂ → H₂O + O₂
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0

Answer:

2H₂O₂ → 2H₂O + O₂

Explanation:

H₂O₂ → H₂O + O₂

The above equation can be balance as illustrated below:

H₂O₂ → H₂O + O₂

There are 2 atoms of O on the left side and a total of 3 atoms on the right side. It can be balance by putting 2 in front of H₂O₂ and 2 in front of H₂O as shown below:

2H₂O₂ → 2H₂O + O₂

Now, the equation is balanced.

**** check ****

Element >>> Left Side >>> Right Side

H >>>>>>>>> 4 >>>>>>>>> 4

O >>>>>>>>> 4 >>>>>>>>> 4

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b) It is based on atomic properties as alkali metals requires 7 more electrons to complete their outer orbit. And they try to give those electrons to other elements to obtain noble gas configuration.

Noble gases are the gases which do not react easily with anything. They are also called as Inert gases, and belongs to group 18 of the periodic table.

Alkali metals are the substances which are found in Group I of a periodic table. Mostly the elements which are present are:

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Properties of alkali metals are:  Soft, shiny reactive metals. They are soft enough to cut with knife. Metals react with water and air quickly and gets tarnish, so pure metals are stored in container by dipping them in oil to prevent oxidation.

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Calculate the molarity of 0.289 moles of FeC13 dissolved in 120 ml of solution
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2.41 M

Explanation:

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A 2.51 times 10^-2 L sample of a solution of Cu^ requires 3.30 times 10^-2 L of 0.132 M KMnO_4 solution to reach the equivalence
sergejj [24]

Answer:

0.868M Cu⁺

Explanation:

The reaction is:

KMnO₄ + 5Cu⁺ + 8H⁺ → 5Cu²⁺ + Mn²⁺ + K⁺ + 4H₂O

<em>Where 1 mol of KMnO₄ reacts with 5 moles of Cu⁺</em>

<em />

3.30x10⁻²L of 0.132M KMnO₄ are:

3.30x10⁻²L ×  (0.132mol KMnO₄ / L) = <em>4.356x10⁻³mol KMnO₄</em>

As 1 mol reacts with 5 moles of Cu⁺, moles of Cu⁺ are:

4.356x10⁻³mol KMnO₄ × (5mol Cu⁺ / 1mol KMnO₄) = <em>0.02178moles Cu⁺</em>

As these moles are in 2.51x10⁻²L, the starting concentration of the Cu⁺ solution is:

<em>0.02178moles Cu⁺ / </em>2.51x10⁻²L = <em>0.868M Cu⁺</em>

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