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Anit [1.1K]
3 years ago
9

What is the pressure of a 59.6-l gas sample containing 3.01 mol of gas at 44.9°c? (r = 0.0821 l • atm/(k • mol), 1 atm = 760 tor

r)?
Physics
1 answer:
defon3 years ago
3 0
<span>Presuming that the gas behaves like an ideal gas: Volume in cubic meter V= 59.6x 0.001= 0.0596 mÂł n=3.01 mol Temperature in Kelvin T=44.9+273.15=318.05 K R= 8.413 J/ mol K The ideal gas law: PxV= nxRxT Substituting the give information from the question into the ideal gas law: P= 3.01 mol x 8.413 J/mol k x 318.05 K / 0.0596= 133544.392 Pa 1351.3458 Pa / 101325 Pa/ atm = 1.3179806 atm</span>
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After all that you have learned in this unit, construct a pamphlet (brochure) in Microsoft Publisher helping new freshman to enc
aleksandr82 [10.1K]
You're not going to like this answer, but it's the only one possible:. It wasn't I who learned anything in this unit. If it was either of us, it was YOU. I can't even tell from reading the question what the topic of the unit was. Was it pamphlets ? Microsoft Publisher ? Freshmen ? Getting Through High School ? This is a lot like asking me to write something "in your own words".
5 0
2 years ago
Question 2 (ID=81813)
zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

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7 0
2 years ago
Read 2 more answers
A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block
gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

  • Mass of first block; m_1 = 5.0kg
  • Mass of second block, m_2 =10kg
  • Tension on first rope; T_1 =\ ?
  • Tension on second rope; T_2 =\ ?

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

F = m\ *\ a

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity  g = 9.8m/s^2 )

For the First Rope

Total mass hanging on it; m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg

So Tension of the rope;

F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

4 0
2 years ago
Read 2 more answers
Please answer both questions and not just one. Thanks!
Margarita [4]
It would be d and c hoped i helped!
8 0
3 years ago
How many nanoseconds does it take light to travel 3.50 ft in vacuum?
Fiesta28 [93]
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
6 0
3 years ago
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