Answer:
Phosphatidylcholine also known as lecithin
Explanation:
The most common phospholipids in order of abundance are
1). phosphatidylcholine, lecithin
2). phosphatidylethanolamine, cephalins 3).phosphatidylinositol, and 4).phosphatidylserine.
They have the common characteristics of esterified fatty acids to the 1 and 2 positions of the glycerol structure with the esterified phosphate group to the 3 position
Answer:
A = Molarity = 0.22 M
B = Molarity = 0.36 M
Explanation:
Given data:
For first solution:
number of moles = 0.550 mol
Volume of solution = 2.50 L
Molarity = ?
Molarity:
Formula:
Molarity = number of moles of solute / volume of solution in L.
Molarity = 0.550 mol / 2.50 L
Molarity = 0.22 M
For second solution:
Mass of NaCl = 15.7 g
Volume of solution = 709 mL or 709/1000 = 0.709 L
Molarity = ?
Solution:
Number of moles = mass / molar mass
Number of moles = 14.7 g/ 58.44 g/mol
Number of moles = 0.252 mol
Molarity:
Molarity = number of moles of solute / volume of solution in L.
Molarity = 0.252 mol / 0.709 L
Molarity = 0.36 M
Answer:
Lose two electrons.
Explanation:
Barium is present in group 2.
It is alkaline earth metal.
Its atomic number is 56.
Its electronic configuration is Ba₅₆ = [Xe] 6s².
In order to attain the noble gas electronic configuration it must loses its two valance electrons.
When barium loses it two electron its electronic configuration will equal to the Xenon.
The atomic number of xenon is 54 so barium must loses two electrons to becomes equal to the xenon.
Answer:
Kc = 168.0749
Explanation:
initial mol: 0.822 0 0
equil. mol: 2(0.822 - x) x x
∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )
⇒ 1.644 - 2x = 0.055 * 1.11
⇒ 1.644 = 2x + 0.06105
⇒ 2x = 1.583
⇒ x = 0.7915 mol equilibrium
⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq
⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²
⇒ Kc = ( 0.7130² ) / ( 0.055² )
⇒ Kc = 168.0749
