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Stells [14]
4 years ago
6

In the periodic table, the elements are organized into groups based on

Chemistry
1 answer:
Pani-rosa [81]4 years ago
8 0
<span>In the periodic table, the elements are organized into groups based on putting together elements with similar properties. For instance, elements in each group have the same number of valence electrons, which makes them form similar bonds. Additionally, elements in the same similar characteristics, such as malleability and magnetism.</span>
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How many sulfur dioxide molecules are there in 1.80 mol of sulfur dioxide
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<span>10.84 to the 23 power</span>
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The following data was collected for the formation of ammonia (NH3) based on the following overall reaction: N2 + 3H2 = 2NH3 N2
notsponge [240]

Answer :  The unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

Explanation :

Rate law 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.

For the given chemical equation:

N_2+3H_2\rightarrow 2NH_3

Rate law expression for the reaction:

\text{Rate}=k[N_2]^a[H_2]^b

where,

a = order with respect to N_2

b = order with respect to H_2

Expression for rate law for first observation:

0.0021=k(0.10)^a(0.10)^b ....(1)

Expression for rate law for second observation:

0.0084=k(0.10)^a(0.20)^b ....(2)

Expression for rate law for third observation:

0.0672=k(0.20)^a(0.40)^b ....(3)

Dividing 2 by 1, we get:

\frac{0.0084}{0.0021}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\4=2^b\\b=2

Dividing 3 by 1 and also put value of b, we get:

\frac{0.0672}{0.0021}=\frac{k(0.20)^a(0.40)^2}{k(0.10)^a(0.10)^2}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[N_2]^a[H_2]^b

\text{Rate}=k[N_2]^1[H_2]^2

Now, calculating the value of 'k' by using any expression.

0.0021=k(0.10)^1(0.10)^2

k=2.1M^{-2}min^{-1}

The value of the rate constant 'k' for this reaction is 2.1M^{-2}min^{-1}

That means, the unit for the rate constant in the rate law for the formation of ammonia is, M^{-2}min^{-1}

7 0
4 years ago
Compound 2 contains 2.0g of hydrogen and 32.0g oxygen. What is the percent compound of each element?
Dmitriy789 [7]

Note down the formula below

\boxed{\sf Mass\%\;of\; element=\dfrac{Mass\:of\:the\: element}{Mass\;of\:the\: compound}\times 100}

Mass of the compound

\\ \sf\longmapsto 32+2=34g

Mass % of Hydrogen:-

\\ \sf\longmapsto \dfrac{2}{34}\times 100

\\ \sf\longmapsto \dfrac{1}{17}\times 100

\\ \sf\longmapsto 5.8\%

Mass % of Oxygen:-

\\ \sf\longmapsto \dfrac{32}{34}\times 100

\\ \sf\longmapsto \dfrac{16}{17}\times 100

\\ \sf\longmapsto 94.2\%

7 0
3 years ago
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