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Lelechka [254]
3 years ago
13

Based on the chart, determine the identity of a substance that increases in temperature by 11.4oC when 1250J of energy are added

to a 55g sample.
Substance Specific Heat (J/g oC)
Lithium 3.56
Vegetable Oil 1.99
Air 1.02
Iron 0.444
Sand 0.290
Gold 0.129

Salt

Air

Lithium

Vegetable Oil
Chemistry
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

LIthium vegetable oil

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Answer:

8.354 nanometers

Explanation:

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\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})

Where Cs is the concentration in the surface of the solid

Cx is the concentration at certain deep X

Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964

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Then we solve for x:

x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm

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alisha [4.7K]
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Tamiku [17]

Explanation:143

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