<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!
Explanation:
Mg+O>MgO
Here the mg loses it electron and is oxidised and oxygen gains and is reduced.
Mg(2+)and O(2-)
Mg is a reducing agent it makes oxygen to be reduced while itself being oxidised and vice versa.
Answer:
Two Half-lives
Explanation:
Let number of Parent nuclei Initially present be X,
Then, finally 
Parent nuclei Will remain with 
daughter nuclei.
In one half- life , parent nuclei becomes half of initial.
So, starting with X parent nuclei,
After one half-life, it will degrade to 
.
After another half life , Parent nuclei will become half of
Which is equal to .
So, Parent nuclei have to go through Two half-lives.
