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ExtremeBDS [4]
2 years ago
11

Which state of matter consists of particles that can be partially compressed? Gas Liquid Plasma Solid

Chemistry
1 answer:
son4ous [18]2 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

In liquid state, particles do have kinetic energy that helps in partially overcoming the intermolecular forces between the molecules. But still the particles are close together and they are able to slide past each other.

So, when we apply pressure on a liquid then its molecules partially gets compressed.

On the other hand, molecules of a solid are held together by strong intermolecular forces of attraction. Hence, they have definite shape and volume. As a result, solids do not get compressed.

In gases and plasma state of matter, molecules are gar away from each other. So, they are able to get completely compressed when a pressure is applied.

Thus, we can conclude that liquid is the state of matter which consists of particles that can be partially compressed.

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Firdavs [7]
Well, the cause of a volcanic eruption is when the lower crust melts. This leads to the formation of magma. The volcano is an opening through which the magma is discharged. The rock inside the earth melts. This magma is diffused upwards through the volcano. If magma reaches the top of the volcano, it behaviour depends on viscosity. So what happens before a volcanic eruption is that the lower crust melts, causing the formation of magma, which is released through the volcano. Hope i helped. 
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El aguacate al oxidarse, ¿que fenómeno es?, ¿físico, químico o alotropicos?​
devlian [24]

Explanation:

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2 years ago
4) The initial rate of the reaction between substances P and Q was measured in a series of
ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

rate = k[P]^{2} [Q]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

k=\frac{rate}{[P]^{2}[Q] } \\  =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4

Second row:

2. Rate value:

rate =0.4* (0.10)^{2} * (0.10)\\\\        =4.0*10^-3mol.dm^-3.s^-1

3.Third row:

[Q]=\frac{rate}{k.[P]^{2} } \\     =9.6*10^-3 / (0.4 *(0.40)^{2} \\    =0.15mol.dm^{-3}

4. Fourth row:

[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}

6 0
2 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

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