1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yulyashka [42]
3 years ago
11

Except for elements 1 and 2, all other elements are stable with how many electrons in their outermost (valence) energy level? ex

cept for elements 1 and 2, all other elements are stable with how many electrons in their outermost (valence) energy level? 2 8 18 32
Chemistry
2 answers:
Elodia [21]3 years ago
6 0
<span>Having 8 electrons in their outermost level of energy, all these elements are stable. The number of electrons in the outermost level is simply called as valance. These are the electrons in the atom that interact and form the chemical bonds. The stability of the element is also determined by the valance.</span>
Artyom0805 [142]3 years ago
4 0
Answer is: <span>other elements are stable with 8 electrons.
For example sodium has eleven electrons and lost one electron to obtain electron configuration of noble gas neon. </span>₁₁Na⁺ 1s²2s²2p⁶.
Chlorine has 17 electrons (seven valence electrons), chlorine gain one electron to obtain electron configuration of noble gas argon. ₁₇Cl⁻ 1s²2s²2p⁶3s²3p⁶.
You might be interested in
Write the Henderson-Hasselbalch equation for a solution of formic acid. Calculate the quotient [HCO2]/[HCO2H] at (a) pH 3.000; (
Elena L [17]

Answer:

a. 0.182

b. 1.009

c. 1.819

Explanation:

Henderson-Hasselbach equation is:

pH = pKa + log [salt / acid]

Let's replace the formula by the given values.

a. 3 = 3.74 + log [salt / acid]

3 - 3.74 = log [salt / acid]

-0.74 = log [salt / acid]

10⁻⁰'⁷⁴ = 0.182

b. 3.744 = 3.74 + log [salt / acid]

3.744 - 3.74 = log [salt / acid]

0.004 = log [salt / acid]

10⁰'⁰⁰⁴ = 1.009

c. 4 = 3.74 + log [salt / acid]

4 - 3.74 = log [salt / acid]

0.26 = log [salt / acid]

10⁰'²⁶ = 1.819

3 0
3 years ago
Balance the following equation in acidic conditions. Phases are optional. Cu NO3,
Gwar [14]
If your equation is <span>Cu+NO^- 3-->Cu^2+NO, then the answer is 
</span><span>2 Cu + 1 NO3{-}  → 1 Cu^{2+} + 3 NO 
</span>
To check if it is balance, this is the solution:
2- Cu- 2
3- N -3
3- O -3 

5 0
3 years ago
The smallest unit of a compound is called a ___________.
Oxana [17]
<span>The smallest unit of a compound is called a molecule. The correct option among all the options that are given in the question is the second option or the penultimate option or option "B". The other choices are incorrrect and can be negated. I hope that this is the answer that has actually come to your desired help.</span>
5 0
3 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
4 years ago
Given the standard heats of reaction
ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s)  → M X₄ (g ) - 98.7 kJ -  2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) +  M (g ) =  M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond =  773.4 / 4 =  193.4 kJ / mole

6 0
3 years ago
Other questions:
  • Please help me!!!! Thank you so much! I’m giving away 25 points for this one! I need both the problems solved
    15·1 answer
  • What is the competition and what are the types of symbiosis?
    6·2 answers
  • Arrange the following elements in order of decreasing atomic radius: Ba, Sn, S, Pb, and As. Rank elements from largest to smalle
    14·1 answer
  • Is H-3 stable?<br> Is He-3 stable?<br> Is He-5 stable?<br> Is Li-7 stable?<br> Is Be-7 stable?
    11·1 answer
  • Balance the combustion reaction between butane and oxygen. 2C4H10 + O2 → CO2 + H2O
    8·2 answers
  • Natalie observes these characteristics in the model of an atom.
    15·1 answer
  • In general, atomic radii:
    15·1 answer
  • Calculate the amount of solute excess out when the 60 gm of saturated solution of
    7·1 answer
  • Part A. Two containers, one at 305 K and the other at 295 K, are placed in contact with each other. 1. 1 J of heat flows from th
    14·1 answer
  • Explain why nitrogen diffuses faster than clorine
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!