The alkali metals all have

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Ag(s) and Ni2+(aq) to give Ni(s) and A

rusak2 [61]

Answer: $3\times 10^{35}$

Explanation:

The balanced chemical equation will be:

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

Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Ag^{+}/Mg]}=+0.80V$

$E^0_{[Ni^{2+}/Ni]}=-0.25V$

$E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}$

$E^0=-0.25-(+0.80V)=-1.05V$

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

$\Delta G$ = gibbs free energy

n= no of electrons gained or lost =?

F= faraday's constant

$E^0$ = standard emf

$\Delta G=-2\times 96500\times (-1.05)=202650J$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin = $25^0C=25+273=298K$

K = equilibrium constant

$\Delta G=-2.303RTlog K$

$+202650=-2.303\times 8.314\times 298\times logK$

$K=3\times 10^{35}$

Thus the value of the equilibrium constant at $25^0C$ is $3\times 10^{35}$

6 0

3 years ago

A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, eac

Andrei [34K]

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) = 0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

<h3>0.432 drinks are toxic</h3>

5 0

3 years ago

What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O

Ipatiy [6.2K]

Answer:

Mass = 13.23 g

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

O₂ : C₃H₈

5 : 1

1.5 : 1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g

6 0

3 years ago

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

6 0

3 years ago

How much heat is required to increase the temperature of 10.0 grams of water 6.0oC? (The specific heat of water is 4.18 J/g x oC

scoray [572]

Heat required in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC (T2-T1)
Heat = 10.0 g (4.18 J/g-C ) ( 6.0 C )
<span>Heat = 250.8 J</span></span>

8 0

3 years ago