The new concentration of the solution is 0.016 M
<h3>Dilution</h3>
From the question, we are to determine the concentration of the new solution prepared from question 4.
NOTE: Question is provided below
From question 4, the concentration of the sodium chloride solution is 0.200M
Now, to determine the new concentration,
Using the dilution law
C₁V₁ = C₂V₂
Where
C₁ is the initial concentration
V₁ is the initial volume
C₂ if the final concentration
and V₂ is the final volume
Then,
C₁ = 0.200 M
V₁ = 10.00 mL
C₂ = ?
V₂ = 125.00 mL
Then,
0.200 × 10.00 = C₂ × 125.00
C₂ = 0.016 M
Hence, the new concentration of the solution is 0.016 M
Question 4:
Calculate the amount of solid sodium chloride needed to make 200.00 mL of a 0.200 M solution
Learn more on Dilution here: brainly.com/question/24881505
Answer:
Light changes speed in a glass of water.
Explanation:
I'm pretty sure the answer is B double check to make sure it isn't A
Explanation:
Alkali metal cations reacts vigorously with halogens to form ionic/electrovalent compounds.
- The alkali metals are about the most reactive metals and they belong to group I .
- In their outermost shell, they have one valence shell electron.
- Alkali metals are largely electropositive willing to release their valence electron to attain stability.
- The halogens have seven electrons in their outermost shell. To complete their configuration, they need just one electron.
- All the halogens are strongly electronegative and they are readily reactive.
- In the vicinity of group 1 elements, they react vigorously.
- This is due to the electrostatic attraction between the metal and non-metal ion formed.
- Therefore, they form strong ionic bonds between their compounds.
Learn more:
sodium alkali and halogens brainly.com/question/6324347
#learnwithBrainly
Answer:
Fluorine has seven electrons in 2p-subshell whereas chlorine has seven electrons in its 3p-subshell. 3p-subshell is relatively larger than 2p-subshell. Therefore, repulsion among the electrons will be more in the 2p-shell of fluorine than 3p-subshell in chlorine. Due to the smaller size and thus, the greater electron-electron repulsions, fluorine will not accept an incoming electron with the same as chlorine.
