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labwork [276]
3 years ago
5

___________ would albumin expect to dissociate into individual ions and increase the number of osmols/mol in solution.

Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer: Check explanation please.

Explanation:

Before diving into the question,let us consider some important facts and definitions.

WHAT IS OSMOLARITY? Osmolarity is the number of OSMOLES of solute per litre of the solution. The unit of osmolary is OSMOLE. Nowadays, osmolarity has become a term used in the past, now osmolarity is called osmotic concentration.

OSMOLARITY can be calculated using; Σ(osmotic coefficient × number of ion × the solute molar concentration).

BACK TO THE QUESTION; ALBUMIN is a PROTEIN. Protein is a large molecule and it contributes an insignificant amount to osmolarity(around 0.6 Mosmole/litre).

Also, if we have the osmotic coefficient to be between zero and one, there is going to be great DISSOCIATION.

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Write all of the allowed sét of quantum numbers for a 3p orbital
V125BC [204]

Answer:

Which of the following is a possible set of quantum numbers for a 3p orbital?

Since you are dealing with a 3p-orbital, your principal and angular momentum quantum numbers will be n=3 and l=1 .

4 0
3 years ago
A 35.6 mL solution of Sr(OH)2 is titrated with 0.549 L of 0.0445 M HBr. Determine the concentration of strontium hydroxide.
ozzi

Answer:

The answer to your question is  0.34 M  

Explanation:

Data

[Sr(OH)₂] = ?

Volume of Sr(OH)₂ = 35.6 ml or 0.0356 l

[HBr] = 0.0445 M

Volume of HBr = 0.549 l

Balanced chemical reaction

                 2HBr  +  Sr(OH)₂  ⇒  SrBr₂   +   2H₂O

Process

1.- Calculate the moles of HBr

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.0445 x 0.549

          = 0.0244

2.- Calculate the moles of Sr(OH)₂ using the coefficients of the balanced reaction

                    2 moles of HBr ----------------------- 1 mol of Sr(OH)₂

                    0.0244 moles   -----------------------  x

                                   x = (0.0244 x 1) / 2

                                   x = 0.0122 moles of Sr(OH)₂

3.- Calculate the concentration of Sr(OH)₂

Molarity = 0.0122/ 0.0356

-Simplification

Molarity = 0.34                      

5 0
3 years ago
What is missing from this graph?<br><br> A. axis labels<br> B. a trendline<br> C. a title
Iteru [2.4K]

Answer:

Axis Labels

Explanation:

The axis labels are usually located on the x and y axis. This graph however is missing those.

hope this helps!

7 0
3 years ago
What is the volume of calcium in 50 grams
lara [203]

Answer:

(volume = mass/density).

Explanation:

what is the density ?

6 0
2 years ago
Determine the concentrations of MgCl2, Mg2+, and Cl− in a solution prepared by dissolving 2.52 × 10−4 g MgCl2 in 2.00 L of water
dybincka [34]

Answer:

Explanation:

Given parameters:

Mass of MgCl₂ = 2.52 x 10⁻⁴g

Volume of water = 2L

Unkown:

Concentration of MgCl₂ =?

Concentration of Mg²⁺ = ?

Concentration of Cl⁻ =?

Solution:

  Concentration is defined as the number of moles of a solute contained in a solution.

   Concentration = \frac{number of moles }{volume}

 To find the number of moles"

           number of moles = \frac{mass}{molar mass}

   Molar mass of MgCl₂ = 24.3 + (2 x 35.5) = 95.3g/mol

       number of moles = \frac{0.000}{molar mass} = 0.000252moles

 Concentration of MgCl₂ = \frac{0.000252}{95.3} = 2.64 x 10⁻⁶moldm⁻³

 

from the formula of the compound;

    1 mole of MgCl₂ contains 1 mole of Mg²⁺

Therefore, 2.64 x 10⁻⁶moldm⁻³ of MgCl₂ will contains 2.64 x 10⁻⁶moldm⁻³ of Mg

Also;

   1 mole of MgCl₂ contains 2 mole of Cl⁻

  2.64 x 10⁻⁶moldm⁻³ contains 2 x 2.64 x 10⁻⁶moldm⁻³; 5.29 x 10⁻⁶moldm⁻³

 

Expressing in ppm;

               1ppm = 1mg/L

   2.64 x 10⁻⁶moldm⁻³ to mg/L for Mg²⁺

2.64 x 10⁻⁶moldm⁻³  = 2.64 x 10⁻⁶moldm⁻³  x molar mass(g/mol)

                                     = 2.64 x 10⁻⁶moldm⁻³ x 24.3 = 6.43 x 10⁻⁵g/L

    g/L to mg/L; 6.43 x 10⁻⁵g/L x 1000 =   6.43 x 10⁻²mg/L = 6.43 x 10⁻²ppm

5.29 x 10⁻⁶moldm⁻³ to mg/L for Cl⁻;

   5.29 x 10⁻⁶moldm⁻³ = 5.29 x 10⁻⁶moldm⁻³ x 35.5 = 1.88 x 10⁻⁴g/L

   g/L to mg/L; 1.88 x 10⁻⁴g/L x 1000 = 1.88 x 10⁻¹mg/L = 1.88 x 10⁻¹ppm

7 0
3 years ago
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