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2 years ago

Question 6 of 25

Chemistry

2 answers:

seraphim [82]2 years ago

6 0

Answer:

thermal

Explanation:

a p e x

Marianna [84]2 years ago

3 0

The answer will be thermal.

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Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

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

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

2 years ago

Assuming that sea water is a 3.5 wt % solution of NaCl in water, calculate its osmotic pressure at 20°C. The density of a 3.5% N

olga nikolaevna [1]

Answer:

π = 14.824 atm

Explanation:

wt % = ( w NaCL / w sea water ) * 100 = 3.5 %

assuming w sea water = 100 g = 0.1 Kg

⇒ w NaCl = 3.5 g

osmotic pressure ( π ):

π = C NaCl * R * T

∴ T = 20 °C + 273 = 293 K

∴ C ≡ mol/L

∴ density sea water = 1.03 Kg/L....from literature

⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln

⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol

⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M

⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K

⇒ π = 14.824 atm

7 0

2 years ago

A pure white crystalline compound was found to melt at 112.5-113.0oC when taken on a melting point apparatus, and on further hea

Sergeeva-Olga [200]

According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.

A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).

However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.

Learn more: brainly.com/question/5325004

8 0

2 years ago

A biochemical reaction will proceed in the direction as written if:_______a. ΔG > 0.b. ΔH > 0.c. ΔG = zero.d. ΔG < 0.e.

garik1379 [7]

Answer:

ΔG < 0

Explanation:

The condition for spontaneity of a chemical reaction is that ∆G<0. Hence for any biochemical reaction to proceed as written, ∆G must have a negative value.

If ∆G >0, the reaction is not spontaneous and will not proceed as written. ∆G=0 means that the reaction has attained equilibrium.

5 0

3 years ago

Some of Earth layers are not solid and are able to flow true or false

aalyn [17]

Answer:

True?

can I have brainiest pls

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7 0

2 years ago