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Rama09 [41]
3 years ago
9

If the heating curve is reversed, what describes the boiling point?

Chemistry
1 answer:
GalinKa [24]3 years ago
5 0
Boiling is the process of converting a substance from liquid state to gaseous state. If the heating curve is reversed, the process also is reversed from converting gaseous state to liquid state. In this case, the reverse of boiling is condensation. So the answer is point of condensation.
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PLEASE HELP!!! 890 g of reactant A completely reacts with 120 g of reactant B to form two products. If product A has a mass of 9
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The following is known as the thermite reaction: 2Al(s)+Fe2O3(s)→Al2O3(s)+2Fe(s) This highly exothermic reaction is used for wel
Colt1911 [192]

Answer:

\Delta H_{rxn}=-847.6\ KJ/mol.

Explanation:

The balanced chemical equation is:

2Al(s)+Fe_2O_3(s)-->Al_2O_3+2Fe(s)

Now, standard values of \Delta H_f\ of\ all\ participants\ of\ reaction.

\Delta H_f(Al_2O_3)=-1669.8\ KJ/mol\\\Delta H_f(Fe_2O_3)=-822.2\ KJ/mol\\\Delta H_f(Al(s))=0\ KJ/mol\\\Delta H_f(Fe(s))=0\ KJ/mol\\

Now, We know \Delta H_{rxn}=\Delta H_{products}-\Delta H_{reactants}

Putting all values of \Delta H_f to above equation.

We get,

\Delta H_{rxn}=-847.6\ KJ/mol.

Hence, this is the required solution.

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3 years ago
Question 2
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How is gravity an attractive force?
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5 0
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Read 2 more answers
What is the vapor pressure of a liquid at 305.03 K if its ∆Hvap = 28.9 kJ/mol and its normal boiling point is 341.88 K?
ASHA 777 [7]

<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm

<u>Explanation:</u>

To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm

P_2 = pressure of the liquid = ?

\Delta H_{vap} = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 341.88 K

T_2 = final temperature = 305.03 K

Putting values in above equation, we get:

\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm

Hence, the vapor pressure of the liquid is 0.293 atm

5 0
3 years ago
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