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xenn [34]
3 years ago
7

Ozone decomposes to oxygen according to the equation 2 O3(g) → 3 O2(g). Write the equation that relates the rate expressions for

this reaction in terms of the disappearance of O3 and the formation of oxygen.
Chemistry
1 answer:
Roman55 [17]3 years ago
7 0

Answer:

Check explanation.

Explanation:

The rate of reaction can be defined as the speed at which concentration of Reactant(s) is/are being converted into product(s) or it can be defined as the rate at which the concentration of Reactant(s) disappear and the rate at which the product(s) forms.

For instance, in the Example below;

aA + bB-------> cC + dD. Where a,b,c and d are the number of moles of A,B,C and D respectively.

The rate of Reaction can be expressed as; - 1/a d[A]/dt = –1/b d[B]/dt= 1/c d[C]/dt =1/d d[D]/dt.

So, back to the question; we are given the balanced equation of,

===> 2 O3(g) --------> 3 O2(g).

Rate of disappearance of ozone,O3 is:

= -∆[O3]/ ∆t.

= -1/2 [O3]/∆t.

Where t= time and the [O3} is the concentration of ozone,O3.

Note that we have the negative sign because, ozone is been consumed/used up in the reaction.

==> The rate of the formation of oxygen is ;

= ∆[O2]/∆t.

= 1/3 ∆[O2]/∆t.

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Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

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An aqueous solution that is 20.0 percent sulfuric acid (h2so4) by mass has a density of 1.105 g/ml. determine the molarity of th
GuDViN [60]

The definition of molarity is:

Molarity = moles of solute / total volume of solution in Liters

In this case, our solute is Sulfuric Acid. To solve this problem, let us assume a basis for calculation.

Let us say that,

Total mass of solution = 100 g

Therefore mass of H2SO4 = 20 g

<span>The molar mass of H2SO4 = 98.079 g / mol</span>

 

Calculating for number of moles of H2SO4:

number of moles = <span>20 g / (98.079 g / mol)</span>

number of moles = 0.204 mol

 

Calculating for the volume of solution using the density:

<span>total volume of solution = 100 g / (1.105 g / mL)</span>

<span>total volume of solution = 90.50 mL = 0.0905 L</span>

 

Therefore the molarity is:

Molarity = 0.204 mol / 0.0905 L

Molarity = 2.2542 mol / L

<span>Molarity = 2.25 M</span>

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