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Vinvika [58]
2 years ago
6

In a cup of hot chocolate, what is the solvent? What is the solute? What is the solution?

Chemistry
2 answers:
Anastaziya [24]2 years ago
3 0
Solvent: The solvent would be the milk (or water, depending on what you used) 
Solute: Chocolate powder
Solution: The hot chocolate itself, since a solution is a homogeneous mixture, in which two or more substances are used

*Forgive me if I am wrong >////<

Hope I helped! <3 
Amanda [17]2 years ago
3 0

Answer:

Solute : Chocolate

Solvent : Milk

Explanation:

<em>Solute </em>- Substance dissolved in the solvent.

<em>Solvent </em>- The substance in which the solute is dissolved.

Since the Chocolate is dissolved in milk, Chocolate is solute and milk is the solvent.

Imagine that there is a cup of milk and you add chocolate to make chocolate milk. Something you add to solvent is solute.

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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
Help please please help please
Anni [7]

Answer:

I don't fully understand what this is about...

Explanation:

sorry :(

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Of these Galilean moons, the Lo moon is very close to Jupiter. The Ganymede moon is the largest of all the Galilean moons, but it is situated very far from Jupiter in comparison to Lo. Thus, the force of attraction between the Lo and Jupiter is very high, it exhibits the greatest gravitational force with Jupiter.  

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