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ANEK [815]
2 years ago
10

For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.50

0 m K2SO4 c. 0.050 m K2SO4 d. 0.500 m NaCl e. 0.500 m C6H12O6 (glucose)
Chemistry
1 answer:
ira [324]2 years ago
7 0

Answer:

The van't hoff factor of 0.500m K₂SO₄ will be highest.

Explanation:

Van't Hoff factor was introduced for better understanding of colligative property of a solution.

By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.

a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.

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You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?
EleoNora [17]
<span>The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to </span>extracellular fluid and equilibrium is reached.
3 0
1 year ago
Read 2 more answers
Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu
katrin2010 [14]

Answer:

T2=276K

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

(V1/T1)=(V2/T2)

T2=( V2*T1)/V1

T2=(322*298)/348

T2=276K

Hence, the temperature of the freezer is 276 K

8 0
2 years ago
In the reaction N2 + 3H2 ⇌ 2NH3, an experiment finds equilibrium concentrations of [N2] = 0.1 M, [H2] = 0.05 M, and [NH3] = 0.00
jasenka [17]
  The   equilibrium  constant  Kc   for  this  reaction    is  calculated  as  follows

from  the  equation   N2  + 3H2 =2 NH3

   qc =   (NH3)2/{(N2)(H2)^3}


Qc   is  therefore  = ( 0.001)2  /{(0.1) (0.05)^3}  = 0.08
3 0
1 year ago
Read 2 more answers
A 251 g strip of glass wool is used to insulate a reaction flask. During the reaction the temperature of the glass wool increase
Ivahew [28]

Answer:

8.9 KJ

Explanation:

Given data:

Mass of strip = 251 g

Initial temperature = 22.8 °C

Final temperature = 75.9 °C

Specific heat  capacity of granite = 0.67 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 75.9 °C - 22.8 °C

ΔT = 53.1 °C

Q = 251 g × 0.67 j/g.°C × 53.1 °C

Q = 8929.8 J

Jolue to KJ.

8929.8J ×1 KJ / 1000 J

8.9 KJ

8 0
1 year ago
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