Solution :
For the reaction :
we have
![$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$](https://tex.z-dn.net/?f=%24Ka%20%3D%20%5Cfrac%7B%5B%5Ctext%7BTris%7D%5E-%20%5Ctimes%20H_3O%5D%7D%7B%5Ctext%7BTris%7D%5E%2B%7D%24)
Clearing 
, we have 
So to reach 
, one must have the 
concentration of the :
![$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$](https://tex.z-dn.net/?f=%24%5Ctext%7B%5BOH%7D%5E-%5D%3D10%5E%7B-pOH%7D%20%3D%206.31%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctext%7B%20moles%20of%20base%7D%24)
So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.
Volume NaOH 
Tris mass 
Now to prepare the said solution we must mix:
gauge to 1000 mL with water.
Answer:
What grade are you in. If i am older, I dont think you would be able to help me.
Explanation:
Answer:
1. Caffeine, C₈H₁₀N₄O₂
Amount = 1.00/194 = 0.00515 moles
2. Ethanol, C₂H₅OH
Amount = 0.0217 moles
3. Dry Ice, CO₂
amount = 0.0227 moles
<em>Note: The question is incomplete. The compound are as follows:</em>
<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>
<em>2. Ethanol, C₂H₅OH;</em>
<em>3. Dry Ice, CO₂</em>
Explanation:
Amount (moles) = mass in grams /molar mass in grams per mole
1. Caffeine, C₈H₁₀N₄O₂
molar mass of caffeine = 194 g/mol
Amount = 1.00 g/194 g/mol = 0.00515 moles
2. Ethanol, C₂H₅OH
molar mass of ethanol = 46 g/mol
Amount = 1.00 g/46 g/mol = 0.0217 moles
3. Dry Ice, CO₂
molar mass of dry ice = 44 g/mol
amount = 1.00 g/44 g/mol = 0.0227 moles
States that the properties of elements are periodic or recurring and are correlated to their atomic number.
Answer:
For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin. Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.
Explanation:
If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.
