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mote1985 [20]
3 years ago
10

Name two device that are using spinning​

Chemistry
1 answer:
ikadub [295]3 years ago
7 0
<h3><u>Two </u><u>device</u><u> </u><u>that </u><u>are </u><u>using</u><u> </u><u>spinning</u></h3>

→ <u>Takli</u>

  • Takli is a small support style spindle. Takli is a perfect tool for spinning cotton, and this are used too to spin fibers, cotton, cashmere, and silk etc. Takli is known as handspindle. And it is the simplest device for spinning.

→ <u>Charkha</u>

  • Charkha is a device for spinning thread or yarn from fibers etc. This are known as spinning wheel.

<h2>HAVE A NICE DAY!</h2>
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

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The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
3 years ago
FIRST ANSWER = BRAINLIEST
STALIN [3.7K]

A_{r} = 24.3

The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.  

We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).  

Thus,  

0.790 × 24 u = 18.96 u

0.100 × 25 u =   2.50 u

0.110 × 26 u =    <u>2.86 u</u>  

       TOTAL =  24.3   u

∴ The relative atomic mass of X is 24.3.


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