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3 years ago

15

Calculate the oxidation numbers of the elements in this equation 2 Al2O3 → 4 Al + 3 O2

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Explanation:

Answer

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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

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2 NO + O22 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the fo

riadik2000 [5.3K]

Answer : The value of rate of reaction is $1.35\times 10^{-8}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given chemical equation is:

$2NO+O_2\rightarrow 2NO_2$

Rate law expression for the reaction is:

$\text{Rate}=k[NO]^a[O_2]^b$

As per question,

a = order with respect to $NO$ = 2

b = order with respect to $O_2$ = 1

Thus, the rate law becomes:

$\text{Rate}=k[NO]^2[O_2]^1$

Now, calculating the value of rate of reaction by using the rate law expression.

Given :

k = rate constant = $9.87\times 10^3M^{-2}s^{-1}$

[NO] = concentration of NO = $7.86\times 10^{-3}M$

$[O_2]$ = concentration of $O_2$ = $2.21\times 10^{-3}M$

Now put all the given values in the above expression, we get:

$\text{Rate}=(9.87\times 10^3M^{-2}s^{-1})\times (7.86\times 10^{-3}M)^2\times (2.21\times 10^{-3}M)^1$

$\text{Rate}=1.35\times 10^{-8}Ms^{-1}$

Hence, the value of rate of reaction is $1.35\times 10^{-8}Ms^{-1}$

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Answer:

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Explanation:

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Products are: CO₂ and H₂O

In this case, the X compound is the solid palmitic acid: C₁₆H₃₂O₂

The balanced equation will be:

C₁₆H₃₂O₂ (s) + 22O₂(g) → 16CO₂(g) + 16H₂O(g)

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