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Nataly [62]
3 years ago
14

I need help asap thank u

Chemistry
1 answer:
UNO [17]3 years ago
4 0
It is b, i learned this last year hope this helps
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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
eduard

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

8 0
2 years ago
Suppose a snack bar is burned in a calorimeter and heats 2,000 g water by 20 °C. How much heat energy was released? (Hint: Use t
Salsk061 [2.6K]

The answer is: 167360

6 0
2 years ago
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A mineral that contains silicon and oxygen
Sergio [31]
<span>Hey there!
Great question:)

Answer:Silicates, this is a mineral that contains silicon and oxygen!

I hope this helps;)
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3 0
3 years ago
A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)
elixir [45]

Answer: \Delta H^{0}=-173.72 kJ/mol

Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}

Ca(OH)_{2}_{(aq)}+2HCl_{(aq)} → CaCl_{2}_{(s)}+2H_{2}O_{(l)}

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: \Delta H^{0}= -1002.82 kJ/mol

Calcium chloride: \Delta H^{0}= -795.8 kJ/mol

Water: \Delta H^{0}= -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]

-17.2=-1367.46+1002.82-2\Delta H

2\Delta H=17.2-364.64

\Delta H=-173.72

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

8 0
2 years ago
126785033% Divided by 348675948849 = <br> what does it equal
meriva

Answer:

0.00000363618

could be wrong.

double check me someone or just trust me

(don't blame me if you get it wrong)

5 0
3 years ago
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