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2 years ago

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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What happens immediately before a volcano erupts?

Firdavs [7]

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3 years ago

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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:

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

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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)