Answer:
high
Explanation:
because of the strong crystal structure mentioned in the question, ionic compounds require a higher level of heat to melt
Answer:
Triacylglycerols (triglycerides) are non-polar or hydrophobic molecules
Glycerophospholipids (phosphoglycerides) are polar or hydrophilic
Explanation:
Triacylglycerides (TGs) are esters synthesized by the esterification of three molecules of glycerol and fatty acid, and this is essentially by the replacement of the three hydroxyl groups on glycerol with three fatty acids. This removes the hydrophilic property from the glycerol molecule hence TGs are only soluble in non-polar solvents like alcohol and benzene, and TGs are therefore the storage forms of fats in adipose tissues.
Glycerophospholipids are formed by the addition of two hydrophobic fatty acid groups and one phosphoric acid (phosphate group) to glycerol (alcohol) leaving a hydrophilic head (phosphate end) and a hydrophobic tail (fatty acid ends). This amphipathic property of this molecule makes it a suitable molecule for membrane structures, especially of cells (lipid bi-layer), with an arrangement where the hydrophilic side interacts with the aqueous environment, while the hydrophobic side makes contact with the non-aqueous environment.
The Arrhenius equation relates activation energy to reaction rates and temperature:
ln (k2 / k1) = (E / R) * (1/T1 - 1/T2), where E is activation energy of 272 kJ, R is the ideal gas constant (we use the units of 0.0083145 kJ/mol-K for consistency, to cancel out the kJ unit), we let T1 = 718 K and k1 = 2.30 x 10^-5, and T2 = 753 K and k2 be the unknown.
ln (k2 / 2.30x10^-5) = (272 kJ / 0.0083145 kJ/mol-K) * (1/718 - 1/753)
k2 = 1.91 x 10^-4 /s
Answer:
pKa = 3.51
Explanation:
The titration of acid solution with NaOH can be illustrated as:
Given that:
Volume of acid solution 
Volume of NaOH 
Molarity of acid solution 
Molarity of NaOH 
For Neutralization reaction:
Making 
the subject of the formula; we have:
However; since the number of moles of NaA formed is equal to the number of moles of NaOH used : Then :
Total Volume after titration = ( 25 + 18.8 ) m
= 43.8 mL
Molarity of salt (NaA ) solution = 
= 
= 0.0429 M
After mixing the two solution ; the volume of half neutralize solution is = 25 mL + 43.8 mL
= 68.8 mL
Molarity of NaA before mixing 
Volume 
Molarity of NaA after mixing 
Volume 
∴
Molarity of acid before mixing = 0.0725 M
Volume = 25 mL
Molarity of acid after mixing = 
= 0.0273 M
Since this is a buffer solution ; then using Henderson Hasselbalch Equation
![pH = pKa + log \frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![3.51= pKa + log \frac{[0.0273]}{[0.0273]} \\ \\ 3.51= pKa + log \ 1 \\ \\ 3.51= pKa + 0 \\ \\ pKa = 3.51](https://tex.z-dn.net/?f=3.51%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5B0.0273%5D%7D%7B%5B0.0273%5D%7D%20%5C%5C%20%20%5C%5C%203.51%3D%20pKa%20%2B%20log%20%5C%201%20%20%5C%5C%20%5C%5C%203.51%3D%20pKa%20%2B%200%20%5C%5C%20%5C%5C%20pKa%20%3D%203.51)
