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iVinArrow [24]
3 years ago
6

A student is testing the effects of temperature on rate of dissolution. She drops food coloring in three different beakers: one

with 100mL of
hot water, one with 100 mL of room temperature water, and one with 100 mL of cold water and predicted that the beaker with the hot

water would spread the fastest. Why do you think she made that prediction?
Chemistry
1 answer:
RSB [31]3 years ago
8 0

Answer: please see anbswer below

Explanation:

She made such prediction because When food colouring  is added into hot water, the difussion ie spreading rate will be fast . This is due to the high kinetic energy of of the molecules which are randomly bouncing against each other causing the solution to mix faster.

When compared with water at room temperature and cold water, the particles are  not moving  as fast and the food coloring  does not spread or diffuse as fast as the hot water because of the lower kinetic energy of molecules in the warm and cold water although the food coloring in warm water will spread faster than in the cold water.

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How many Oxygen atoms in: 5H2SO3
astra-53 [7]

Answer:

20

Explanation: I think

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The rate constant for this second‑order reaction is
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Answer:

daddadaddddddddddddddddddddddddddddda

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3 0
3 years ago
Write balanced equations that describe the following reactions. (Use the lowest possible coefficients. Use the pull-down boxes t
Gnesinka [82]

Answer :  The balanced chemical reaction will be:

HBrO_4(aq)+H_2O(l)\rightarrow H_3O^+(aq)+BrO_4^-(aq)

Explanation:

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

As we know that perbromic acid is considered a strong acid that means it will completely dissociate in water.

The balanced chemical reaction will be:

HBrO_4(aq)+H_2O(l)\rightarrow H_3O^+(aq)+BrO_4^-(aq)

8 0
3 years ago
Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so
Marat540 [252]

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:  

C₆H₅-COOH + HCO₃⁻  ⇄  C₆H₅-COO⁻  +  H₂CO₃

C₆H₅-OH + HCO₃⁻  ⇄  no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.      

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl  ⇄  C₆H₅-COOH + Cl⁻

<u>This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration</u>.  

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻  ⇄  C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl  ⇄  C₆H₅-OH + Cl⁻  

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This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.  

I hope it helps you!

6 0
3 years ago
How many joules of energy do you have if you have 0.45 L of water in a can and you heat the 23C water to 55C site:socratic.org
Bess [88]

Answer:

Q = 60192 j

Explanation:

Given data:

Volume of water = 0.45 L

Initial temperature = 23°C

Final temperature = 55°C

Amount of heat absorbed = ?

Solution:

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 =  55°C - 23°C

ΔT = 32°C

one L = 1000 g

0.45 × 1000 = 450 g

Specific heat capacity of water is 4.18 j/g°C

Q = m.c. ΔT

Q = 450 g.  4.18 j/g°C.  32°C

Q = 60192 j

4 0
3 years ago
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