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Jlenok [28]
3 years ago
11

7.The _____ theory of matter states that all particles of matter are in constant motion.

Chemistry
2 answers:
SashulF [63]3 years ago
6 0
The kinetic energy theory of matter states that all particles of matter are in constant motion.
Kinetics has to do with some kind of movement, which is why this answer is the only plausible one. 
Setler79 [48]3 years ago
4 0

The kinetic theory of matter states that all particles of matter are in constant motion.

The following are the postulates of kinetic theory of gases:

1. The particles of gases are in constant and random motion. Due to this motion, they collide with each other and walls of the container.

2. The particles of gases have point mass and zero volume.

3. The particles of gases have no such attraction or repulsion between them.

4. The kinetic energy with which the particles are moving is directly proportional to temperature. A temperature increases, kinetic energy also increases, and particles collide each other more frequently.

5. The particle of gases at given temperature have same kinetic energy.



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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo
riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)

The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

\Delta H = heat released by the reaction = ?

m = mass of benzene = 125 g

c_{p,g} = specific heat of gaseous benzene = 1.06J/g^oC

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

4 0
2 years ago
Which of these is a characteristic of science? (5 points) Question 1 options: 1) It cannot be reproduced by any scientist. 2) It
Alex17521 [72]

Answer:

3

Explanation:

It is based on empirical evidence

4 0
3 years ago
A drop of water with a mass of 0.48 g is vaporized at 100 ∘C and condenses on the surface of a 55- g block of aluminum that is i
bagirrra123 [75]

The final temperature in Celsius of the metal block is 49°C.

<h3>How to find the number of moles ?</h3>

Moles water = \frac{\text{Given mass}}{\text{Molar Mass}}

                     = \frac{0.48\ g}{18\ \text{g/mol}}

                     = 0.0266 moles  

                   

Heat lost by water = 0.0266 mol x 44.0 kJ/mol

                                = 1.17 kJ

                                = 1170 J           [1 kJ = 1000 J]

Heat lost = Heat gained

Heat gained by aluminum = 1170 J  

1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242  

1170 + 1242 = 49.7 T  

T = 48.5°C (49°C at two significant figures)

Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.

Learn more about the Moles here: brainly.com/question/15356425

#SPJ1

7 0
1 year ago
Can you help me please
irina [24]

Answer:

Concentrated solutions

7 0
2 years ago
Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significa
elena-s [515]

Answer:

1.12 × 10⁻⁴ M

Explanation:

Step 1: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 2: Make an ICE chart

We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.

       Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

I                                0                    0

C                              +S                +2S

E                                S                  2S

The solubility product constant is:

Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³

S = 1.12 × 10⁻⁴ M

8 0
2 years ago
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