Actually since Bromine is located at the 1 Carbon, so we
can say that this is a primary alkyl halide and which undergoes SN2 or E2
reactions. This reaction is a bimolecular, single step process because it is a
primary.
<span>The substitution product formed will be 1-ethoxybutane
(main product) and sodium bromide (side product).</span>
We are given with the balanced equation above 2Na + 2H2O = <span>2NaOH + H2. when 22.4 L of H2 at STP is present, there is a one mole equivalent of H2. Via stoichiometry, there are 2 moles of Na needed. The equivalent mass of Na is equal to 45.98 grams. ANswer is D</span>
The molecular shape : this is trigonal planar
the molecule is : Non polar
Answer:
Explanation:
<u>1) Data:</u>
a) n = 2 moles
b) T = 373 K
c) V = 2.5 liter
d) P = ?
<u>2) Chemical principles and formula</u>
You need to calculate the pressure of the propane gas in the mixture before reacting. So, you can apply the partial pressure principle which states that each gas exerts a pressure as if it occupies the entire volume.
Thus, you just have to use the ideal gas equation: PV = nRT
<u>3) Solution:</u>
P = 2 mol × 0.08206 atm-liter /K-mol × 373K / 2.5 liter = 24.5 atm
Since the number of moles are reported with one significant figure, you must round your answer to one significant figure, and that is 20 atm (20 is closer to 24.5 than to 30).
Answer:
Explanation:
Hello,
In this case, considering the dissociation of thiamine hydrochloride:
It is convenient to write it as:
Or:
With which the Henderson-Hasselbalch equation is applied:
![pH=pKa+log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Therefore:
![log(\frac{[A^-]}{[HA]} )=3.50-[-log(3.37x 10^{-7})]=3.50-6.47=-2.97}\\\\\frac{[A^-]}{[HA]} =10^{-2.97}=1.07x10^{-3}](https://tex.z-dn.net/?f=log%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29%3D3.50-%5B-log%283.37x%2010%5E%7B-7%7D%29%5D%3D3.50-6.47%3D-2.97%7D%5C%5C%5C%5C%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D10%5E%7B-2.97%7D%3D1.07x10%5E%7B-3%7D)
![[A^-]=1.07x10^{-3}}[HA]](https://tex.z-dn.net/?f=%5BA%5E-%5D%3D1.07x10%5E%7B-3%7D%7D%5BHA%5D)
Thus, as the pH equals the concentration of hydrogen which also equals the concentration of the conjugate base, one obtains:
![[H]^+=[A^-]=10^{-pH}=10^{-3.50}=3.16x10^{-4}M](https://tex.z-dn.net/?f=%5BH%5D%5E%2B%3D%5BA%5E-%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-3.50%7D%3D3.16x10%5E%7B-4%7DM)
Now, solving for the concentration of acid ([HA]):
![[HA]=\frac{3.16x10^{-4}M}{1.07x10^{-3}} =0.296M](https://tex.z-dn.net/?f=%5BHA%5D%3D%5Cfrac%7B3.16x10%5E%7B-4%7DM%7D%7B1.07x10%5E%7B-3%7D%7D%20%3D0.296M)
Finally, the mass turns out:
Best regards.
