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slavikrds [6]
3 years ago
9

According to the Clausius theorem, the cyclic integral offor a reversible cycle is zero.​

Chemistry
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

Yes it is true

Explanation:

This is because the Clausius theorem states that The cyclic integral always has two defined results. The results include it being less than or equal to zero under certain conditions.

When the system consists of only reversible processes, the cyclic integral is equal to zero. If it consists of and irreversible processes, the integral is usually less than zero.

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In a certain time, light travels 4.96 km in a vacuum. during the same time, light travels only 3.36 km in a liquid. what is the
Natali5045456 [20]

the refractive index of the liquid is 1.476

The refractive index, which has no dimensions, measures how quickly light passes through a substance.

It can also be described as the difference between the speed of light in a vacuum and a medium.

Refractive index is equal to the product of the light's liquid and vacuum speeds.

Therefore.

speed of light in vacuum = 4.96 km/t

speed of light in liquid = 3.36 km/t

Refractive index = 4.96/3.36

Refractive index =1.476

Therefore, the refractive index of the liquid is 1.476

To learn more about refractive index

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6 0
1 year ago
Estimate the heat of neutralization of H3PO4 in kJ/mol given these data: Heat transferred = 2.5 x 102 J Moles of H3PO4 = 1.4 x 1
OlgaM077 [116]
Below are the steps to get the answers:

<span>1.) write out the balance equation 3NaOh+H3PO4->Na3PO4+3H2O
 
2.) You are given everything needed to calculate q=heat transfer=2.2*10^2, H3PO4 moles= 1.5*10^-3, NaOH moles=5.0*10^-3
 
3.) equation is deltaHneutraliztion=q/Moles of limiting reagent H3PO4 is limiting reagent because lowest moles, and is used up first
 
4.) Now plug in variables DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole Notice we had to convert J to kj, </span>
7 0
3 years ago
Which of these compounds are molecular?<br> CBr4<br> FeS<br> P4O6<br> PbF2
olasank [31]

Answer:

I think it's answer is P4O6

I hope it's helpful for you...

8 0
3 years ago
The Henry's law constant (kH) for O2 in water at 20°C is 1.28 × 10−3 mol/(L·atm). (a) How many grams of O2 will dissolve in 4.00
Burka [1]

Answer:

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>

4 0
2 years ago
In what form does the sun's energy arrive on Earth?The physical characteristics of two large predatory animal species are shown
eimsori [14]

Answer:

radiation B species x because short canines are less likely to break.

6 0
2 years ago
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