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Marysya12 [62]
2 years ago
9

Elements 3 to 10 (Li to Ne) show a more or less steady increase in IE. What does this tell you about the energy level that each

successive electron (from 3 to 10) is assigned to?
Chemistry
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Elements 3 to 10 (Li to Ne) show a more or less steady increase in ionization energy.

<h3>What is ionisation energy?</h3>

The amount of energy required to remove an electron from an isolated atom or molecule.

The major difference is the increasing number of protons in the nucleus as you go from lithium to neon. That causes greater attraction between the nucleus and the electrons and so increases the ionization energies. In fact the increasing nuclear charge also drags the outer electrons in closer to the nucleus.

Learn more about the ionisation energy here:

brainly.com/question/20658080

#SPJ1

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Why wouldn’t a tennis player with two sneakers “bond” to any other tennis players? Why wouldn’t a helium atom with two electrons
aleksklad [387]

Answer:

Explained below.

Explanation:

A tennis player with two sneakers wouldn't bond to any other tennis player because he is already stable and complete with the 2 and doesn't need another players assistance to make him stand well.

However, helium atom with two electrons wouldn't bond to any other atoms because it is stable. This stability arises from the fact that it has two protons and 2 electrons, of which the 2 electrons completely fill its valence shell/outer most shell to make it neutral.

8 0
4 years ago
Determine the empirical formulas for compounds with the following percent compositions:
White raven [17]

Answer: a)  CS_2

b) CH_2O

Explanation:

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 15.8g

Mass of S= 84.2 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{15.8g}{12g/mole}=1.32moles

Moles of S=\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{84.2g}{32g/mole}=2.63moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{1.32}{1.32}=1

For S =\frac{2.63}{1.32}=2

The ratio of C  : S = 1:2

Hence the empirical formula is CS_2

b) Mass of C= 40 g

Mass of H= 6.7 g

Mass of O = 53.3 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.7g}{1g/mole}=6.7moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.33}{3.33}=1

For H = \frac{6.7}{3.33}=2

For O =\frac{3.33}{3.33}=1

The ratio of C : H: O= 1 :2: 1

Hence the empirical formula is CH_2O

5 0
3 years ago
How chemistry is a part of your morning routine
natali 33 [55]
When you wake up early, and go brush your teeth the toothpaste is made of chemical compounds, the water that you use to wash your faceis treated chemically. To make the coffee sugar or sweetener has chemical compounds.

hope this helps!
8 0
3 years ago
Read 2 more answers
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Andreas93 [3]

My number one tip is to STUDY! I know it sucks, but once you're in high school, it's going to become a part of your routine every single day. Even if you study right now, keep on doing it over and over and over. Trust me! It'll help. Also, get a tutor! Maybe someone you know is doing well and you could ask them to help, or you could ask a student who was in your class and passed. My number one advice is to just keep working hard. I get it; chemistry is HARD. You'll get there!

5 0
3 years ago
Read 2 more answers
You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration
vovangra [49]

<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM



7 0
4 years ago
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