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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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Which energy profile best shows that the enthalpy of formation of CS2 is 89.4 KJ/mol?
Anna11 [10]

Answer:

Option C. Energy Profile D

Explanation:

Data obtained from the question include:

Enthalpy change ΔH = 89.4 KJ/mol.

Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:

Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.

If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.

Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.

Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.

7 0
2 years ago
Read 2 more answers
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
2 years ago
One example of a thermal conductor
Harlamova29_29 [7]
Metal radiator. When hot water flows through the coils of the radiator , the metal heats up quickly by conduction and then radiates thermal energy into the surrounding air.


Answer: Metal radiator
6 0
2 years ago
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It is desired to produce 2.25 grams of dichloromethane (CH2Cl2) by the following reaction. If the percent yield of dichlorometha
Allushta [10]

Answer:- 3.12 g carbon tetrachloride are needed.

Solution:- The balanced equation is:

CH_4+CCl_4\rightarrow 2CH_2Cl_2

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.

percent yield formula is:

percent yield = (\frac{actual}{theoretical})100

65.5=(\frac{2.25}{theoretical})100

theoretical=(\frac{2.25(100)}{65.5})

theoretical = 3.44 g

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Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.

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= 3.12gCCl_4

So, 3.12 grams of carbon tetrachloride are needed to be reacted.

8 0
2 years ago
What forms when iron-53 decays?
amm1812
It depends on the type of decay that is taking place if its a β+ it will decay into Mn-52 β- decays into Co-59
3 0
3 years ago
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