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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 piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand throug
djyliett [7]

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>

<em></em>

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>

<em></em>

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial  = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

7 0
3 years ago
Que tipo de mistura tem a agua mineral?
QveST [7]

Answer:

mezcla homogénea

El agua en sí es un ejemplo de mezcla homogénea. Todo el agua, excepto la más pura, contiene minerales y gases disueltos. Estos se disuelven en todo el agua, por lo que la mezcla se presenta en la misma fase y es homogénea.

6 0
3 years ago
WILL GIVE BRAINLIEST PLZ HELP!!!!!!
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Answer:

D

Explanation:

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6 0
3 years ago
Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility.
Sindrei [870]

The molar solubility is 7.4×10^{-3} M and the solubility is  7.4×10^{-3} g/L .

Calculation ,

The dissociation of silver bromide is given as ,

AgBr → Ag ^{+} + Br^{-}

S  

 -          S        S        

Ksp =  [Ag ^{+} ] [ Br^{-} ]  =  [S] [ S ] = S^{2}

S = √ Ksp = √ 5. 5×10^{-5} = 7.4×10^{-3}

The solubility =7.4×10^{-3} g/L

The molar solubility is the solubility of one mole of the substance.

Since ,  one mole of AgBr is dissociates and form one mole of each  Ag ^{+} and Br^{-} ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4×10^{-3} mol/L = 7.4×10^{-3} M

To learn more about molar solubility ,

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7 0
1 year ago
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Answer:

air gas

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candle wax solid or melted would be liquid

Explanation:

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