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alina1380 [7]
3 years ago
9

Which is the electron configuration for nobelium (No)?

Chemistry
2 answers:
marishachu [46]3 years ago
6 0
It’s [Xe]7s^2 5f^14 (first one) hopefully i could help!!
Stolb23 [73]3 years ago
6 0

Answer:

[Rn]7s25f14

Explanation:

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Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d
worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.  

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:

A+B\rightarrow Products

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

3.20\times 10^{-1}=k(1.50)^a(1.50)^b ....(1)

Expression for rate law for second observation:

3.20\times 10^{-1}=k(1.50)^a(2.50)^b ....(2)

Expression for rate law for third observation:

6.40\times 10^{-1}=k(3.00)^a(1.50)^b ....(3)

Dividing 1 from 2, we get:

\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0

Dividing 1 from 3, we get:

\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[A]^1[B]^0

\text{Rate}=k[A]

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

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Explanation:

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