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3 years ago

The diagrams below represent the four states of matter.

Chemistry

2 answers:

UkoKoshka [18]3 years ago

6 0

Diagram 1 exhibits the nature of the particles within the state that forms after a solid melts. Solids melt into liquids and the particles within a liquid have a greater spacing than the particles in a solid. Moreover, these particles are free to slip over one another, which means that liquids do not have a definite shape; however, the particles are still confined in by intermolecular forces, which means that the volume of a liquid is definite.

anastassius [24]3 years ago

3 0

Answer: Diagram 1 (with the blue balls).

Explanation:

1) After the solids melt they are liquids.

Liquid state is characterized by the particles being close but yet free to move, they can slip over each other, which is what permit the liquids flow and take the shape of the containter.

2) Solid is the state of the matter in which the particles are stuck together, they cannot move with respect to the others, but they are in fixed positions. In this state the particles vibrate and the matter has a definite shape and volume.

This state is represented by the diagram 3 (the red balls).

3) In gas state the particles are quite separated, free to move in and occupy all the space. In this state the particles are moving at great speed. The volume of the gas is the volumen of the container.

This state is represented by the diagram 2(the yellow balls).

4) Plasma is the state of the matter similar to gas state but the particles are ions. Those are particles that have charge, either positive or negative.

This is represented by the diagram 4 (the balls with positive and negative symbols).

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3 years ago

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3 years ago

Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa

sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

$k = A e^\frac{-E_{a}}{RT}$

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² = 1,85x10¹²

I hope it helps!

8 0

3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

3 years ago