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miskamm [114]
3 years ago
7

how many moles of water will be produced from the complete combustion of 24.6 moles of ethane according to the following reactio

n
Chemistry
1 answer:
lorasvet [3.4K]3 years ago
7 0
Combustion of ethane:

C2H6 + 7/2 O2 > 2 CO2 + 3 H2O
1 mol of ethane produces 3 mol of water
24.6 mol of ethane produce?mol of water
24.6/3 =8.2 mol of water is produced
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True or False? In a C++ floating-point constant, a decimal point is not required if exponential (E) notation is used.
defon

Answer:

The answer is True

Explanation:

This is because they are equivalent and have same number of bit.

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3 years ago
Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou
Georgia [21]

We know,

\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol

For given reaction, I_2(g)+Cl_2(g)\ -->\ 2ICl(g)

\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol

For , 2.41 moles of I_2 :

\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ

We know :

\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K

Hence, this is the required solution.

7 0
3 years ago
If a gas occupies a volume of 950 mL at standard temperature, what volume will it
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Voulme 1= 950 mL
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Temperature 2 = 50 C
Convert your temperature to Kelvin
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Temperature 1 = 25 C + 273 = 298 K
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Plug in to the Formula
950 mL/298 K = ? / 323 K

Rearrange the formula to make one to solve for what is missing.
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3 0
2 years ago
Which statement describes how heat is transferred by radiation?
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4 0
3 years ago
Read 2 more answers
Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu
nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^0_{[Pb^{2+}/Pb]}=-0.13V

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E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy = ?

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F = Faraday constant = 96500 C/mole

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Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0
2 years ago
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