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Rufina [12.5K]
3 years ago
11

Which is a postulate of the kinetic molecular theory of gases? Despite being very small, the volume occupied by the individual p

articles that compose a gas cannot be assumed to be negligible. The volume occupied by the individual particles that compose a gas is very small. However, the distances between gas particles are comparable to the sizes of the gas particles, so the volume occupied by the individual gas particles cannot be assumed to be negligible. The particles that compose a gas are so small compared to the distances between them that the volume of the individual particles can be assumed to be negligible. Because the distances between gas particles are so small relative to the sizes of the gas particles themselves, the volume of the individual particles can be assumed to be negligible.
Chemistry
2 answers:
Paladinen [302]3 years ago
7 0

Answer:

The particles that compose a gas are so small compared to the distances between them that the volume of the individual particles can be assumed to be negligible.  

Explanation:

This is a postulate of the Kinetic Molecular Theory.

A is wrong. KMT assumes the that the volume of the particles is negligible.

B is wrong. KMT assumes that the distance between the particles is muck greater than their size.

D is wrong. It takes the large distances as a fact. KMT uses this as an assumption.

Darina [25.2K]3 years ago
4 0

Answer:

Gas particles have a small volume relative to the spaces between them.

Explanation:

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calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o
Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of Na_3PO_4 = 12.00 g

Mass of CaCl_2 = 10.0 g

Molar mass of Na_3PO_4 = 164 g/mol

Molar mass of CaCl_2 = 111 g/mol

Molar mass of NaCl = 58.5 g/mol

Molar mass of Ca_3(PO_4)_2 = 310 g/mol

First we have to calculate the moles of Na_3PO_4 and CaCl_2.

\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}

\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol

and,

\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}

\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2

From the balanced reaction we conclude that

As, 3 mole of CaCl_2 react with 2 mole of Na_3PO_4

So, 0.0901 moles of CaCl_2 react with \frac{2}{3}\times 0.0901=0.0601 moles of Na_3PO_4

From this we conclude that, Na_3PO_4 is an excess reagent because the given moles are greater than the required moles and CaCl_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl  and Ca_3(PO_4)_2

From the reaction, we conclude that

As, 3 mole of CaCl_2 react to give 6 mole of NaCl

So, 0.0901 mole of CaCl_2 react to give \frac{6}{3}\times 0.0901=0.1802 mole of NaCl

and,

As, 3 mole of CaCl_2 react to give 1 mole of Ca_3(PO_4)_2

So, 0.0901 mole of CaCl_2 react to give \frac{1}{3}\times 0.0901=0.030 mole of Ca_3(PO_4)_2

Now we have to calculate the mass of NaCl  and Ca_3(PO_4)_2

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g

and,

\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2

\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

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Neporo4naja [7]
Okay thanks for the update I will give you a call when you get home thanks
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