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Vedmedyk [2.9K]
2 years ago
10

Immunoglobulin G (IgG), formerly called gamma globulin, is a principal antibody in blood serum. A 0.470 g sample of immunoglobul

in G is dissolved in water to make 0.106 L of solution, and the osmotic pressure of the solution at 25 ∘C is found to be 0.733 mbar. Calculate the molecular mass of immunoglobulin G.
Chemistry
1 answer:
Nookie1986 [14]2 years ago
5 0

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:

brainly.com/question/5925156

#SPJ1

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Anton [14]

Answer:

presence or absence of a nucleus

Explanation:

These classification of organisms into broad domains is based on the present or absence of nucleus in the cell of an organism.

The archaea are prokaryotes and they lack a distinct cellular nuclei.

Bacteria are similar to archaea but bacteria have only one RNA polymerase

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So, the variations in their cells are used to classify organisms into the broad categories

8 0
3 years ago
Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa
Mademuasel [1]

Answer : The vapor pressure of propane at 25.0^oC is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of propane at 25.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 25.0^oC=273+25.0=298.0K

T_2 = normal boiling point of propane = -42.04^oC=230.96K

\Delta H_{vap} = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})

P_1=17.73atm

Hence, the vapor pressure of propane at 25.0^oC is 17.73 atm.

3 0
3 years ago
Jessica has 12.7 moles of a compound for an experiment. How many particles of the compound does she have?
guajiro [1.7K]

Answer:

Number of moles is defined as the ratio of given mass in g to the molar mass.The mathematical expression is given as:

Number of moles  =Number of moles of compound  = 12.7 moles (given)As, 1 mole of any compound is equal to  particles.

where,    is Avogadro number. Formula for calculating particles is given by:

where, N =  number of particles, n = number of moles and  is Avogadro number.

Put the values,=  or  Hence, number of particles of the compound is equal to

Explanation:

8 0
3 years ago
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Brums [2.3K]
I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
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