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3 years ago

A species whose oxidation number increases in a reaction is

1 answer:

7 0

A species whose oxidation number increases in a reaction is oxidized (reducing reagents or reactant).
For example, balanced chemical reaction: 2KClO₃ → 2KCl + 3O₂.
Oxygen in the reactant state has oxidation number -2 (+1 + 5 + 3 · x = 0) and it is oxidized to 0 (in molecule of oxygen) in the product state.

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Identify the correct coefificients to balance it -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to:

$5$ $\rm O_2$ molecules.

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

5 0

2 years ago

How is burning magnesium different than burning methane

olya-2409 [2.1K]

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

brainly.com/question/13824679

#learnwithBrainly

6 0

2 years ago

What describes a change that occurs when one substance reacts to another substance?

Slav-nsk [51]

A chemical or physical change

8 0

3 years ago

Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for ho

nikdorinn [45]

To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]

HOBr = 0.50 M
KOBr = 0.30 M = OBr-

 HOBr + H2O <-> H+ + OBr- 
I 0.50 - 0 0.30 
C -x x x
---------------------------------------------
E(0.50-x) x (0.30+x) 

Assuming that the value of x is small as compared to 0.30 and 0.50 

Ka = 2.0 x 10^-9 = x (0.30) / 0.50) 

x = 3.33 x 10^-9 = H+
pH = 8.48

3 0

3 years ago

1. A 10 g gold coin is heated from 25°C to 50°C (CAu is 0.13 J/g-°C). What is the H?

Oksanka [162]

The correct answer is + 32.5 J because heat is absorbed here by copper not evolved also after calculation we will find the H value is positive, lets calculate it:
H = m C Δt where:
m = 10 g
C = 0.13 J/g.°C
Δ t = final temperature - initial temperature = 50 - 25 = + 25 °C
so H = 10 x 0.13 x (+25) = + 32.5 J

4 0

3 years ago