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BabaBlast [244]
3 years ago
7

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

Chemistry
1 answer:
Alex3 years ago
7 0
Answer is: d) 7.
Electron configuration: 1s²2s²2p⁵.
<span>Valence electrons are in 2s and 2p orbitals.
</span>Valence electrons<span> are outer shell </span>electrons (transition metal can have a valence electrons in an inner shell) <span>that is associated with an </span>atom <span>and they can form </span>chemical bond. V<span>alence eletrons determine chemical proporties of element.</span>
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Type the correct answer in the box.
Murljashka [212]

Answer:

<em>a= In scientific notation</em>

6.96×10⁵ Km

<em>b =In  expanded notation</em>

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

<em>In scientific notation</em>

6.96×10⁵ Km

Size of bacterial cell:

1.9 ×10⁻⁴ mm

<em>In  expanded notation</em>

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<span>+, and OH-</span>?

b) Calculate <span>Kb</span> for ephedrine.

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