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LenKa [72]
3 years ago
14

Using knowledge of intermolecular forces, explain why the yellow food coloring mixed with the water and the acetone was able to

remove the blue dye from the glitter.
(Basically the experiment was we mixed yellow food coloring with acetone and water and added it to glitter, and it removed the blue dye from the glitter. Please mention intermolecular forces in your answer.)
Chemistry
1 answer:
Papessa [141]3 years ago
8 0

The yellow coloring is a polar substance hence it dissolve in water. Blue dye is nonpolar hence it dissolves in less polar acetone.

In chemistry, there is a general rule that like dissolves like. A substance can only dissolve in another if there is some kind of intermolecular interaction between the substances.

The yellow food coloring is a polar substance hence it is able to dissolve in water which is a polar solvent. The blue dye is a nonpolar substance hence it dissolve in acetone which is a less polar solvent.

Learn more:  brainly.com/question/9847214

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How many of protien consumed by humans around the world is fish?
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3 0
4 years ago
If you purchased 0.610 μCi of sulfur-35, how many disintegrations per second does the sample undergo when it is brand new?Expres
ASHA 777 [7]

Answer:

the sample undergo 1.67 \times 10^5  disintegrations per second.

Explanation:

Given:

The amount of sulphur purchased is 0.671 μCi

To find:

disintegrations per second = ?

Solution:

Some of the conversions are

1 Ci = 3.8 \times 10^{10} Bq

1 rad = 0.01 Gy

1Gy = 1 j/kg tissue

1 rem = 0.01 Sv

1Sv = 1 j/Kg

Using these conversions,

The decay rate of this sample is calculated as

0.671 \mu C i \times \frac{1 \mu C i}{10^{6}} \mu C i \times \frac{0.671 \times 10^{10}}{1 \mu C i}

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7 0
4 years ago
Using the appropriate ksp value from appendix d in the textbook, calculate the ph of a saturated solution of ca(oh)2 .
Alex777 [14]
The ph of a saturated solution of Ca(OH)2 is 12.35

CALCULATION:
For the reaction 
     Ca(OH)2 → Ca2+ + 2OH- 
we will use the Ksp expression to solve for the concentration [OH-] and then use the acid base concepts to get the pH:
     Ksp = [Ca2+][OH-]^2

The listed Ksp value is 5.5 x 10^-6. Substituting this to the Ksp expression, we have
     Ksp = 5.5 x 10^-6 = (s) (2s)^2 = 4s^3
     s3 = 5.5x10^-6 / 4

Taking the cube root, we now have
     s = cube root of (5.5x10^-6 / 4)s
        = 0.01112

We know that the value of [OH-] is actually equal to 2s:
     [OH-] = 2s = 2 * 0.01112 = 0.02224 M

We can now calculate for pOH:          
     pOH = - log [OH-]
              = -log(0.02224)
              = 1.65

Therefore, the pH is
     pH = 14 - pOH
           = 14 - 1.65
           = 12.35
4 0
4 years ago
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