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omeli [17]
3 years ago
10

The atoms and molecules in a gas are in constant motion. Temperature is a measure of this motion. How are these related? As temp

erature increases, the molecules and atoms move faster. As temperature decreases, the molecules and atoms move farther apart. As temperature increases, the atoms and molecules move closer together. As temperature decreases, the molecules and atoms move faster.
Chemistry
1 answer:
Sedaia [141]3 years ago
8 0

Here we have to choose the correct statement on the effect of temperature on the motion of the molecules and atoms of a gas.

As the temperature increases the molecules and atoms move faster.

As per the kinetic theory of gas molecules and atoms the kinetic energy (K.E.) of the atom or molecules is related to temperature by the equation

K.E. = \frac{3}{2}kT ( k = Boltzmann constant, T = temperature.

Thus as the temperature increases the K.E. increases thus the atom or molecules move faster.

With the decrease of temperature the movement of the atoms or molecules will be less and they will be near to each other.

The increment of temperature increase the K.E. thus the atoms or molecules move apart from each other.

With the decrease of temperature the movement of the atoms or molecules decreases.  

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Answer:

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Explanation:

Electrolysis of acidified water

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8 0
2 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
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Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

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