What are TWO control variables?

If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give

tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

The maximum height attained by nitrogen gas molecule = h
The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

$K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J$

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

The potential energy at height h = $P.E = 6.381\times 10^{-21} J$
Molar mass of nitrogen gas = 28.0134 g/mol
Mass of nitrogen gas molecule = m

$m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg$

The acceleration due to gravity = g = 9.8 m/s^2
The maximum height attained by nitrogen gas molecule = h
The potential energy is given by:

$P.E=mgh$

$6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km$

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Learn more about the average kinetic energy of gas particles here:

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6 0

2 years ago

WS Percent yield don’t understand how to do would appreciate the help

Grace [21]

Answer:

1. Theoretical yield of NaOH is 22.72 g

2. Percentage yield of NaOH = 22.14%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaHCO₃ —> NaOH + CO₂

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.

Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.

Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e

Xmol of NaHCO₃ = 25 / 44.01

Xmol of NaHCO₃ = 0.568 mole

1. Determination of the theoretical yield of NaOH.

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.

Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.

Thus, the theoretical yield of NaOH is 22.72 g

2. Determination of the percentage yield of NaOH.

Theoretical yield of NaOH = 22.72 g

Actual yield of NaOH = 5.03 g

Percentage yield of NaOH =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.03 / 22.72 × 100

Percentage yield of NaOH = 22.14%

4 0

3 years ago

If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?

QveST [7]

So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K

4 0

3 years ago

If the air in a room has lower temperature than the air coming out of a floor vent from a forced air heater , heat would flow

jolli1 [7]

To the top. Common knowledge my dude. Heat always rises to the top.

5 0

3 years ago

Read 2 more answers

A bond between a nonmetal and a nonmetal is called a(n)

Pani-rosa [81]

Answer:

answer: B