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AfilCa [17]
4 years ago
15

You observe a vaporous gas coming off the top of a boiling pot of water. which of the following is a true statement related to t

his observation?
A: you are observing a chemical property of the water.

B: you are observing a physical property of the water.

C: the water has undergone oxidation.

D: the water must be part of a solution.
Chemistry
1 answer:
Ede4ka [16]4 years ago
6 0

the answer will be A because its a chemical its changing to gas

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Which statement is true about a catalyst?
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Answer:

The answer is D. It changes the final products of the reaction. .

Explanation:

  • <u>. Catalyst is a substance that increases the rate of a reaction without itself being consumed.</u>
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What is interpolation?
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Explanation:<3 sorry.

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The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of th
kotykmax [81]

Answer:

\Delta _{fus}H=3255.3J/mol

\Delta _{fus}S=7.62\frac{J}{mol*K}

Explanation:

Hello,

Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )

Solving for the enthalpy of fusion we obtain:

\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol

Finally the entropy of fusion is given by:

\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}

Best regards.

5 0
3 years ago
2Al(s)+Fe2O3(s)−→−heatAl2O3(s)+2Fe(l) 2Al(s)+Fe2O3(s)→heatAl2O3(s)+2Fe(l) If 26.1 kg Al26.1 kg Al reacts with an excess of Fe2O3
Dmitry_Shevchenko [17]

Answer : The mass of Al_2O_3 produced will be, 49.32 Kg

Explanation : Given,

Mass of Al = 26.1 Kg  = 26100 g

Molar mass of Al = 26.98 g/mole

Molar mass of Al_2O_3 = 32 g/mole

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{26100g}{26.98g/mole}=967.38moles

Now we have to calculate the moles of Al_2O_3.

The balanced chemical reaction is,

2Al+Fe_2O_3\rightarrow Heat+Al_2O_3+2Fe

From the balanced reaction we conclude that

As, 2 moles of Al react to give 1 mole of Al_2O_3

So, 967.38 moles of Al react to give \frac{967.38}{2}=483.69 moles of Al_2O_3

Now we have to calculate the mass of Al_2O_3.

\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3\times \text{Molar mass of }Al_2O_3

\text{Mass of }Al_2O_3=(483.69mole)\times (101.96g/mole)=49317.0324g=49.32Kg

Therefore, the mass of Al_2O_3 produced will be, 49.32 Kg

3 0
3 years ago
Explain why the compound Cl2O has a low melting and boiling point.​
Dima020 [189]
It has a low melting and boiling point as it is unstable and the bonds can break easily.
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