Answer:
Explanation:
The half-cell reduction potentials are
Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = 0.7996 V
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) E° = -0.447 V
To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.
The anode is the electrode at which oxidation occurs.
The equation for the oxidation half-reaction is
Answer:
252.68 K or -20.46 °C
Explanation:
According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".
Mathematically,
P₁ / T₁ = P₂ / T₂ ---- (1)
Data Given:
P₁ = 30.7 kPa
T₁ = 0.00 °C = 273.15 K
P₂ = 28.4 kPa
T₂ = <u>???</u>
Solving equation for T₂,
T₂ = P₂ T₁ / P₁
Putting values,
T₂ = 28.4 kPa × 273.15 K / 30.7 kPa
T₂ = 252.68 K or -20.46 °C
61.24 is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 degrees celcius and 744.0 mm Hg.
Explanation:
given that:
density = 0.00249 g/ml (
) or 2.49 grams/litre
P = 744 mm Hg OR 0.978 atm
T = 20 Degrees or 293.15 Kelvin
R = 0.08206 Litre atm/mole K
molar mass =?
Formula used/
PV = nRT equation 1
here n is number of moles:
n = 
putting the value of n and value of density in the equation 1:
PV = 
x RT
molar mass = x 
= density x
= 
= 61.24 is the molar mass of the gas.
Answer:
Rb2CO3(aq)+Fe(C2H3O2)2(aq)--> 2Rb(C2H3O2)(aq) + FeCO3(s)
Explanation:
The reaction shown in the answer is the reaction of rubidium carbonate and iron II acetate. Rubidium is far more reducing than Fe II hence it can displace Fe II from its salt as shown.
The reducing property of metals depends on the value of their individual electrode potential values. For rubidium, its standard reduction potential is -2.98 V while that of Fe II is -0.44V. Hence rubidium can displace Fe II from its salt as shown above.
