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3 years ago

15

In an energy diagram, the reactants are at a higher potential energy compared to the products. Which of the following best descr

ibes this reaction?

1 answer:

tino4ka555 [31]3 years ago

7 0

It is exothermic and will have a negative enthalpy.

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Which element of the periodic table is named after the moon

Troyanec [42]

Answer:

The answer to your question is Selenium

Explanation:

The origin of the names of the elements comes from different origins

For example

Sanskrit words. 14 elements have roots from this language

Planets, 4 elements took their names from the planets.

Greek, 42 elements took their names from these languages

The name of Selenium comes from the greek that means moon.

8 0

3 years ago

HELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPP

guajiro [1.7K]

Hiii,so the problem is ‍♀️

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2 years ago

Can aromatic compounds belong to other classes of compounds?

Elena L [17]

They certainly can. However, they have other groups that are used to classify a compound.

8 0

3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

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

where,

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

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago

Consider the reaction: <img src="https://tex.z-dn.net/?f=NiO%28s%29%2BCO%28g%29%20%5Crightleftharpoons%20Ni%28s%29%2BCO_2%28g%29

sasho [114]

Answer:

Molar concentration of CO₂ in equilibrium is 0.17996M

Explanation:

Based on the reaction:

NiO(s) + CO(g) ⇆ Ni(s) + CO₂(g)

kc is defined as:

kc = [CO₂] / [CO] = 4.0x10³ <em>(1)</em>

As initial concentration of CO is 0.18M, the concentrations in equilibrium are:

[CO] = 0.18000M - x

[CO₂] = x

Replacing in (1):

4.0x10³ = x / (0.18000-x)

720 - 4000x = x

720 = 4001x

x = 0.17996

Thus, concentrations in equilibrium are:

[CO] = 0.18000M - 0.17996 = 4.0x10⁻⁵

[CO₂] = x = <em>0.17996M</em>

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Thus, <em>molar concentration of CO₂ in equilibrium is 0.17996M</em>

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I hope it helps!

5 0

3 years ago