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Andrews [41]
4 years ago
13

Example. A Student wants to increase the maximum if sugar that can dissolve in water..She crushes the sugar and then stirs it in

to the water .Does This work?
Chemistry
2 answers:
Advocard [28]4 years ago
8 0

Answer:

Yes this works.

Explanation:

This works because when the sugar is crushed, more of the surface area of the sugar gets exposed to the solvent, allowing the solvent to dissolve it a lot quicker than when it wasn't crushed.

Stirring also helps the sugar particles dissolve faster. This is because the particles are now agitated, which increases their kinetic energy.  This increase in Kinetic energy is reflected by increase in temperature, which helps the particles dissolve faster.

Delvig [45]4 years ago
6 0

Answer: yes it does work.

Explanation:

When the sugar is crushed by the student, an increase in the number of molecules of sugar occurs. Meaning that more number of the solute (sugar molecules) are able to interact with the solvent (the water).

Therefore, there will be more number of collisions taking place between the solute and solvent molecules.

Hence, we can say it works because this leads to more interaction or collisions between the solute and solvent particles, as a result of this increased interaction between solute and the solvent, sugar will dissolve readily into the water.

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The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan
BlackZzzverrR [31]

Answer:

The equilibrium constant at 2000 K is 0.7139

The equilibrium constant at 3000 K is 8.306

ΔH = 122.2 kJ/mol

Explanation:

Step 1: Data given

the standard molar Gibbs energy of formation of X(g) is 5.61 kJ/mol at 2000 K

the standard molar Gibbs energy of formation of X(g) is -52.80 kJ/mol at 3000 K

Step 2: The equation

1/2X2(g)⟶X(g)

Step 3: Determine K at 2000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 2000 K

⇒K is the equilibrium constant

5610 J/mol = -8.314 J/molK * 2000 * ln K

ln K = -0.337

K = e^-0.337

K = 0.7139

The equilibrium constant at 2000 K is 0.7139

Step 4: Determine K at 3000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 3000 K

⇒K is the equilibrium constant

-52800 J/mol = -8.314 J/molK * 3000 * ln K

ln K = 2.117

K = e^2.117

K = 8.306

The equilibrium constant at 3000 K is 8.306

Step 5: Determine the value of ΔH∘rxn

ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)

ln 8.306 /0.713 = -ΔH/8.314 * (1/3000 - 1/2000)

2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)

2.455 = -ΔH/8.314 * (-1.67*10^-4)

-14700= -ΔH/8.314

-ΔH = -122200 J/mol

ΔH = 122.2 kJ/mol

6 0
4 years ago
How many moles of carbon are there in 5 g of carbon?
Umnica [9.8K]

Answer:

0.416666667

Explanation:

number of moles= mass of sample ÷ molar mass

=5÷12

=0.41666667

8 0
3 years ago
How many molecules of water are in a 45 g sample of H2O?
Ymorist [56]

Answer:

3. 0.75 mole. or (C)

Explanation:

1.5 moles. 9.03X1023 . 3. 0.75 mole. 4581023. 4. 15 moles.

4 0
3 years ago
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What do you mean tell me one thing you learned what did you learn
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