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3 years ago

7

How many grams of aluminum are needed to react completely with 1.2 mol of FeO?

1 answer:

a_sh-v [17]3 years ago

4 0

Answer:

5

Explanation:

all you do is the math expression

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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

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

3 years ago

PLEASE HELP DUE IN EXACTLY 15 mins!! i will give you branliest

Deffense [45]

$2032533 \sq22222rt[2]{?}$

4 0

3 years ago

Which of the following is the correct isotope notation for the element above?

ozzi

7 becuse it splits in half

6 0

3 years ago

Read 2 more answers

The molarity of concentrated acetic acid is 17.45 M. When acetic acid is diluted in water it is commonly called vinegar with a m

Sladkaya [172]

M1 = 17.45 M

M2 = 0.83 M

V2 = 250 ml

M1. V1= M2. V2

V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml

7 0

3 years ago