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
Answer: The pH of a 4.4 M solution of boric acid is 4.3
Explanation:
at t=0 cM 0 0
at eqm 
So dissociation constant will be:
Give c= 4.4 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D4.4%5Ctimes%200.000011%3D4.8%5Ctimes%2010%5E%7B-5%7DM)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[4.8\times 10^{-5}]=4.3](https://tex.z-dn.net/?f=pH%3D-log%5B4.8%5Ctimes%2010%5E%7B-5%7D%5D%3D4.3)
Thus pH of a 4.4 M 
solution is 4.3
Q = mcΔθ
67.5 = m x 0.45 x (28.5 - 21.5)
M = 67.5 / 3.15
= 21.4 g
