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atroni [7]
3 years ago
8

Which one is correct for number 6 I can’t decide

Chemistry
1 answer:
nadezda [96]3 years ago
5 0

I'm pretty sure it's D .

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The vapor pressure of dichloromethane, c h 2 c l 2 , at 0 ∘ c is 134 mmhg . the normal boiling point of dichloromethane is 40. ∘
sladkih [1.3K]
<span> the </span>vapor pressure<span> of the liquid at a temperature T</span>2<span> ... Now, </span>it's<span> important to realize that the </span>normal boiling point<span> of a substance is measured at an atmoshperic ... ΔHvap=−ln(</span>134mmHg760mmHg<span> )⋅8.314J mol−1K−1 (1(273.15+</span>0)−1(273.15+40))K−1 ... Give equations that can be used tocalculate<span> the .

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3 years ago
What is electrons configuration for arsenic ?
Anna35 [415]

Answer:

[Ar] 3d¹⁰ 4s² 4p³

Explanation:

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2 years ago
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Fudgin [204]

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The answer is A. Chemical

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What are the weighting laws?
lora16 [44]

Answer: A comprehensive legal term for uniform standards ascribed to the quantity, capacity, volume, or dimensions of anything. Legislation that adopts and mandates the use of a uniform system of weights and measures is a valid exercise of Police Power, and such laws are constitutional.

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8 0
2 years ago
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

E^0_{[Cu^{2+}/Cu]}=+0.34V

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 = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

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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