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Levart [38]
2 years ago
11

We might think of a porous material as being a composite wherein one of the phases is a pore phase. Estimate upper and lower lim

its for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air.
Chemistry
1 answer:
eduard2 years ago
8 0

Answer:

The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are

Ku = 38.252 W/mK

K lower = 0.199 W/mK

Explanation:

As we know  

Ku = Vp * Kair + Vmagnesium * K metal  

Ku = 0.10 *0.02 + (1-0.25) * 51

Ku = 38.252 W/mK

The lower limit  

K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)

K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)

K lower = 0.199 W/mK

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Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

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Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

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Explanation:

To calculate the moles, we use the equation:

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\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

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According to stoichiometry :

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Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

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