The three major groups on the peridoic table are the ___

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

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

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

A 1.67-g sample of solid silver reacted in excess chlorine gas to give a2.21-g sample of pure solid Agcl.The heat given off in t

kotegsom [21]

Given:

Mass of Ag = 1.67 g

Mass of Cl = 2.21 g

Heat evolved = 1.96 kJ

To determine:

The enthalpy of formation of AgCl(s)

Explanation:

The reaction is:

2Ag(s) + Cl2(g) → 2AgCl(s)

Calculate the moles of Ag and Cl from the given masses

Atomic mass of Ag = 108 g/mol

# moles of Ag = 1.67/108 = 0.0155 moles

Atomic mass of Cl = 35 g/mol

# moles of Cl = 2.21/35 = 0.0631 moles

Since moles of Ag << moles of Cl, silver is the limiting reagent.

Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles

Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol

Ans: Formation enthalpy = 126.5 kJ/mol

6 0

3 years ago

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The town in the video stressed using locally made goods instead of imported goods. How could using locally made goods lower CO2

ladessa [460]

Answer:

It reduces the need to import goods

Explanation:

When you buy locally, the products you buy don't come from far away, so they don't have to cross the country (or the ocean) by boat, plane or trucks to reach the market/store where you're buying, at least not from a long distance away.

The distance a vehicle travels, the less CO2 emissions it produces.

If the good you're buying is made/produced only an hour away, that's not much pollution produced compared as if the good has to come from a distant place spending days on highways to reach you.

6 0

3 years ago

a solution of sodium hydroxide, 0.0500 M, is used to titrate a 15.00 mL sample of hydrochloric acid to the endpoint. The initial

Akimi4 [234]

Answer:

The answer to your question is Molarity = 0.0708

Explanation:

Data

NaOH 0.05 M Volume 1 = 3.87 ml Volume 2 = 25.11 ml

HCl 15 ml

Process

1.- Find the volume used of NaOH

25.11 - 3.87 = 21.24 ml = 0.02124 l

2.- Write the balanced equation of the reaction

NaOH + HCl ⇒ NaCl + H₂O

3.- Calculate the moles of NaOH in the solution

Molarity = $\frac{moles}{volume}$

moles = Molarity x volume

moles = 0.05 x 0.02124

moles = 0.001062

4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.

1 mol of NaOH ------------- 1 mol of HCl

0.001062 moles of NaOH ------------ x

x = (0.001062 x 1) / 1

x = 0.001062 moles of HCl

5.- Find the molarity of HCl

Molarity = $\frac{0.001062}{0.015}$

Molarity = 0.0708

4 0

3 years ago

What is density? What does it tell you about the substance?

olya-2409 [2.1K]

How heavy the object is.

7 0

3 years ago

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The three major groups on the peridoic table are the ____