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Oksana_A [137]
3 years ago
6

The vapor pressure of methanol is 143 mmhg. identify the best reason to explain why methanol spontaneously evaporates in open ai

r at 25.0∘c and standard pressure (760 mmhg). the vapor pressure of methanol is 143 . identify the best reason to explain why methanol spontaneously evaporates in open air at 25.0 and standard pressure (760 ).
Chemistry
1 answer:
goldfiish [28.3K]3 years ago
6 0

Answer/Explanation:

Methanol has a molecular weight (32.04 g/mol), low-boiling point and because of its low boiling point, methanol readily evaporates at room temperature.

Under these specified non-standard conditions, the partial pressure of methanol is lower than its vapor pressure and this explains the reason for the spontaneous evaporation exhibited by methanol.

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How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
Blast furnaces extract pure iron from the iron(III) oxide in iron ore in a two step sequence.In the first step, carbon and oxyge
Alona [7]

Answer:

The net chemical equation for the production of iron from carbon, oxygen and iron(III) oxide:

2Fe_2O_3 (s) +6C (s) + 3O_2 (g)\rightarrow 4Fe (s) + 6CO_2 (g)

Explanation:

Step 1: Carbon and oxygen react to form carbon monoxide:

2C (s) + O_2 (g)\rightarrow 2CO (g)..[1]

Step 2: Iron(III) oxide and carbon monoxide react to form iron and carbon dioxide:

Fe_2O_3 (s) + 3CO (g)\rightarrow 2Fe (s) + 3CO_2 (g)..[2]

The net chemical equation for the production of iron from carbon, oxygen and iron(III) oxide can be obtained by :

3 × [1] + 2 × [2]

2Fe_2O_3 (s) + 6CO (g)+6C (s) + 3O_2 (g)\rightarrow 4Fe (s) + 6CO_2 (g)+6CO(g)

CO gas will gas cancel out form both sides ans we will get the net chemical equation:

2Fe_2O_3 (s) +6C (s) + 3O_2 (g)\rightarrow 4Fe (s) + 6CO_2 (g)

According to reaction, 2 moles of ferric oxide reacts with 6 moles of carbon and 3 moles of oxygen gas to give 4 moles of iron and 6 moles pf carbon dioxide gas.

8 0
4 years ago
If the accepted value for the mass of an object is 20.0g and the student found that the mass of the object was 20.5g what is the
NikAS [45]

Answer:

Results

The percent error between 20 and 20.5 is 2.5%

Explanation:

Percent Error = | (20.5 − 20) / 20 | × 100 = | (0.5) / 20 | × 100 = | 0.025 | × 100 = 2.5% (three decimal places)Percent Error = 2.5%

3 0
3 years ago
Is cotton opaque or not
qaws [65]
Cotton is opaque.  Opaque objects do not let light pass through.  You can't see through the objects that are opaque.
3 0
3 years ago
If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after t
Naya [18.7K]
Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
6 0
4 years ago
Read 2 more answers
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