Answer:
Allows diffusion of sub...(true)
its found in both anim...(true)
produces en...(false)
Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Initial concentration = ?
Final concentration = 12.5 grams
Applying in the above equation, we get that:-
![12.5\ g=[A_0]e^{-0.02297\times 90.6}](https://tex.z-dn.net/?f=12.5%5C%20g%3D%5BA_0%5De%5E%7B-0.02297%5Ctimes%2090.6%7D)
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
Answer:
3.62x10⁻⁷ = Kb
Explanation:
The acid equilibrium of a weak acid, HX, is:
HX + H₂O ⇄ X⁻ + H₃O⁺
Where Ka = [X⁻] [H₃O⁺] / [HX]
And basic equilibrium of the conjugate base, is:
X⁻ + H₂O ⇄ OH⁻ + HX
Where Kb = [OH⁻] [HX] / [X⁻]
To convert Ka to Kb we must use water equilibrium:
2H₂O ⇄ H₃O⁺ + OH⁻
Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]
Thus, we can obtain:
Kw = Ka*Kb
Solving for Kb:
Kw / Ka = Kb
1x10⁻¹⁴ / 2.76x10⁻⁸ =
3.62x10⁻⁷ = Kb
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
Answer:
hypothesis
Explanation:
a question that can be answered wity a hypothesis is a question a scientist can answer
