Answer:
12 moles of F₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e
Xmol of F₂ = (3 × 8)/2
Xmol of F₂ = 12 moles
Thus, 12 moles of F₂ is needed for the reaction.
Answer:
See explanation below
Explanation:
In picture 1 attached, we have the starting structure. In picture 2, you have the answer for this.
Basically this is an acid base reaction to promove then, a condensation of the molecule.
In the first step, the carbonile substract a H atom from the HCl. Then, by resonance the OH+, pass the charge to the carbon, which is tertiary and can be easily stabilized. Now, after this, the Chlorine substract an Hydrogen to get a double bond on the carbon with the charge, and this be stable. Now, a second molecule of benzaldehyde reacts with this, so we can have a condensation product.
See picture 2 for complete mechanism.
Answer:
C is the answer they are looking for but clearly the truth is D
Explanation:
Answer:
The correct answer is option B.
Explanation:
Michaelis–Menten 's equation:
![v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D%3Dk_%7Bcat%7D%5BE_o%5D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![V_{max}=k_{cat}[E_o]](https://tex.z-dn.net/?f=V_%7Bmax%7D%3Dk_%7Bcat%7D%5BE_o%5D)
v = rate of formation of products
[S] = Concatenation of substrate = ?
= Michaelis constant
= Maximum rate achieved
= Catalytic rate of the system
= initial concentration of enzyme
We have :

[S] =?

![v=V_{max}\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7Bmax%7D%7D%7B4%7D%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%280.0050%20M%2B%5BS%5D%29%7D)
![[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M](https://tex.z-dn.net/?f=%5BS%5D%3D%5Cfrac%7B0.005%20M%7D%7B3%7D%3D1.7%5Ctimes%2010%5E%7B-3%7D%20M)
So, the correct answer is option B.
Adenine.
Thymine always pairs with Adenine while Cytosine always pairs with Guanine on DNA strands.
Hope this helps!