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Stella [2.4K]
3 years ago
7

What is the average kinetic energy of a gas at 285 kelvin? (R = 8.314 J/K-mol)

Chemistry
2 answers:
maxonik [38]3 years ago
6 0
The answer would be 3.55 x 10^3 j/mol
Svetllana [295]3 years ago
6 0

Answer:

The average KE = 312 J/mol

Explanation:

<u>Given:</u>

Temperature of the gas, T = 285 K

Gas constant, R = 8.314 J/K-mol

<u>To determine:</u>

The average kinetic energy of the gas

<u>Explanation:</u>

Based on the kinetic theory of gases, the average kinetic energy is given as:

KE = \frac{3}{2} RT\\

In this case:

KE = \frac{3}{2} *8.314\ J/K-mol * 285\ K = 312 J/mol

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A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm.
Fed [463]

Answer:

Explanation:

This problem is very similar to the other that you put before, so, we'll use the same principle here.

The ideal gas equation: PV = nRT

Where:

P: pressure in atm

V: volume in L

T: Temperature in K.

n: moles

R: Gas constant (In this case, we'll use 0.082 L atm/K)

to get the molar mass of the gas, we need to know the moles, and with the mass, we can know the molar mass. However we can put the ideal gas expression with the molar mass in this way:

we know that n is mole so:

n = g/MM

If we put this in the idea gases expression we have:

PV = gRT/MM

Solving for MM we have:

MM = gRT/PV

Now, let's convert the temperature and volume to K and L respectively:

T = 67 + 273 = 340 K

V = 350 / 1000 = 0.35 L

Now all we have to do is put all the data into the expression:

MM = 0.79 * 0.082 * 340 / 0.9 * 0.35

MM = 22.0252 / 0.315 = 69.92 g/mol rounded 70 g/mol

Now, the closest answer of your options would be 72 g/mol. This could be easily explained because we do not use all the significant figures of all numbers, including the gas constant of R. However, all the work, calculations and procedure is correct and fine, and we only have a minimum range of 2 units.

6 0
3 years ago
The center on your school’s basketball team is 6 ft 10 in tall. How tall is the player in millimeters (mm)?
svp [43]

The answer is 6 ft 10 inches in millimeters (mm) is 0.833 ft.

Given,

The center of the school's basketball team is 6 ft 10 inches tall.

We have to convert the height of the player from feet and inches to feet.

Using the conversion factor,

1 ft = 12 inches

or, 12inches/ 1 ft

Converting 6ft 10 inches to ft, we get;

10 inches × 1 ft/ 12inches

= 0.833 ft

Therefore 6 ft 10 inches in millimeters (mm) is 0.833 ft.

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

To learn more about Millimeter and Unit conversions, visit: brainly.com/question/26371870

#SPJ4

5 0
2 years ago
How many moles of chloride ions are there in 2.5 L of 5 M magnesium chloride?
castortr0y [4]

Answer:

n_{Cl^-}=25molCl^-

Explanation:

Hello,

In this case, since the given 5-M concentration of magnesium chloride is expressed as:

5\frac{molMgCl_2}{L}

We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:

n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-

Best regards.

4 0
3 years ago
A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not
AleksAgata [21]

Volume of tank = 6.9 m^{3} (given)

Since, 1 m = 3.28 ft

So,  

1 m^{3} = 35.287 ft^{3}

For 6.9 m^{3}:

6.9\times 35.287 ft^{3} = 243.4803 ft^{3}

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

243.4803 ft^{3} \simeq 243.5 ft^{3}

Hence, volume in ft^{3} is 243.5.

5 1
3 years ago
Read 2 more answers
PLEASE HELP ME<br><br> what would be the mass of 9.03 x 1021 molecules of hydrobromic acid (HBr)?
Nikitich [7]

We know 1 mole of any atom or molecules contains 6.033 \times 10^{23} atom or molecules.

1 mole of HBr i.e 81 gm/mol contains  6.033 \times 10^{23} atom or molecules.

So, mass of 9.03\times 10^{21} molecules is :

m=\dfrac{81\times 9.03\times 10^{21} }{6.033 \times 10^{23}}\\\\m= 1.21\ gm

Therefore, mass of  9.03\times 10^{21} molecules is 1.21 gm .

Hence, this is the required solution.

7 0
3 years ago
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