The molecular mass of the immunoglobulin G, given the data from the question is 1.53×10⁵ g/mole
<h3>How to determine the molarity</h3>
We'll begin by calculating the molarity of the immunoglobulin G. This is illustrated below:
- Volume = 0.106 L
- Temperature (T) = 25 °C = 25 + 273 = 298 K
- Osmotic pressure (π) = 0.733 mbar = 0.733 × 0.000987 = 0.00072 atm
- Gas constant (R) = 0.0821 atm.L/Kmol
- Van't Hoff factor (i) = 1
- Molarity (M)
π = iMRT
M = π / iRT
M = 0.00072 / (1 × 0.0821 × 298)
M = 0.000029 M
<h3>How to determine the mole of immunoglobulin G</h3>
- Molarity = 0.000029 M
- Volume = 0.106 L
- Mole =?
Mole = Molarity × volume
Mole = 0.000029 × 0.106
Mole = 3.074×10⁻⁶ mole
<h3>How to determine the molar mass of mmunoglobulin G</h3>
- Mole = 3.074×10⁻⁶ mole
- Mass = 0.470 g
- Molar mass =?
Molar mass = mass / mole
Molar mass = 0.47 / 3.074×10⁻⁶
Molar mass = 1.53×10⁵ g/mole
Learn more about Osmotic pressure:
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The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide.
Balancing the combustion reaction,
C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
(12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
(12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g
Answer:
It is C. observed using the 5 senses.
Explanation:
you can observe a substance without changing its identity.
