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3 years ago

How many valence electrons are represented in the following electron configuration? 1s22s22p5

1 answer:

Alex3 years ago

7 0

Answer is: d) 7.
Electron configuration: 1s²2s²2p⁵.
Valence electrons are in 2s and 2p orbitals.
Valence electrons are outer shell electrons (transition metal can have a valence electrons in an inner shell) that is associated with an atom and they can form chemical bond. Valence eletrons determine chemical proporties of element.

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Type the correct answer in the box.

Murljashka [212]

Answer:

a= In scientific notation

6.96×10⁵ Km

b =In expanded notation

0.00019 mm

Explanation:

Given data:

Radius of sun = 696000 Km

size of bacterial cell = 1.9 ×10⁻⁴ mm

Radius of sun in scientific notation = ?

Size of bacterial cell in expanded notation = ?

Solution:

Radius of sun:

696000 Km

In scientific notation

6.96×10⁵ Km

Size of bacterial cell:

1.9 ×10⁻⁴ mm

In expanded notation

1.9/ 10000 = 0.00019 mm

8 0

3 years ago

Based on the name, which substance is a covalent compound?

V125BC [204]

QUICK ANSWER

The name of the covalent compound N2O5 is dinitrogen pentoxide, more commonly known as nitrogen pentoxide. This covalent compound is part of a bigger group of compounds, nitrogen oxides, created purely from nitrogen and oxygen

6 0

3 years ago

Read 2 more answers

Ephedrine, a central nervous system stimulant, is used in nasalsprays as a decongestant. this compound is a weak organic base: {

sashaice [31]

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-?

b) Calculate Kb for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH⁻] = 10∧(-2,67) = 0,00213 M.
[OH⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.
[C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.
Kb = [OH⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].
Kb = (0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

3 0

3 years ago

What is the pressure of 5.0 mol nitrogen (N2) gas in a 2.0 L container at 268

CaHeK987 [17]

Answer: 54.94atm

Explanation: Please see attachment for explanation

8 0

3 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago