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3 years ago

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4. How does the kinetic theory of gases explain the weather changes happening in the troposphere? You can also research on the I

nternet using keywords such as troposphere, atmospheric layers, and weather changes.

2 answers:

nirvana33 [79]3 years ago

7 0

the kinetic theory shows how energy is transferred throughout the troposphere and how it works.

uranmaximum [27]3 years ago

4 0

According to the kmt pressure is directly proportional to the number of collision between particles

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A sample of an ideal gas at 1.00 atm and a volume of 1.87 L was placed in a weighted balloon and dropped into the ocean. As the

Ludmilka [50]

Answer:

Volume of sample after droping into the ocean=0.0234L

Explanation:

As given in the question that gas is idealso we can use ideal gas equation to solve this;

Assuming that temperature is constant;

Lets $P_1$ and $V_1$ are the initial gas parameter before dropping into the ocean

and $P_2$ and $V_2$ are the final gas parameter after dropping into the ocean

according to boyle 's law pressure is inversly proportional to the volume at constant temperature.

hence,

$P_1V_1=P_2V_2$

P1=1 atm

V1=1.87L

P2=80atm

V2=?

After putting all values we get;

V2=0.0234L

Volume of sample after droping into the ocean=0.0234L

7 0

3 years ago

If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams

Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of $KCl_3$ = 2.0 moles

Molar mass of $O_2$ = 32 g/mole

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 2.0 moles of $KClO_3$ react to give $\frac{2.0}{2}\times 3=3.0$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g$

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}$

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = $214.0^oC=273+214.0=487K$

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

$\rho$ = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

$P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}$

$P=1.78atm$

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0

3 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

3 0

2 years ago

What is the expected hybridization of the central atom tetrahedral

Pepsi [2]

Answer:

Tetrahedral molecules are normally spy hybridized.

Explanation:

4 0

2 years ago

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Even though the same amount of energy was transferred into both substances from the air, the isopropanol evaporated while the wa

Nesterboy [21]

Answer:

The isopropanol evaporated while the water did not because the molecules don't stick together as strongly as the molecules in the water do. The water would need more energy transferred in, in order to evaporate.

Explanation:

4 0

3 years ago