Answer:
The first one
Explanation:
Elements are as simple as it gets
it is most likely a beaker, beacuse 300ml is quite a large volume. Otherwise, it would be a measuring cylinder or pippette
<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
Answer:
A. 2
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