Answer: i would say producer
Explanation:
First, we need to get the number of moles:
from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+
∴ the number of moles n = 4
we are going to use this formula:
㏑K = n *F *E/RT
when K is the equilibrium constant = 4.98 x 10^-5
and F is Faraday's constant = 96500
and the constant R = 8.314
and T is the temperature in Kelvin = 298 K
and n is number of moles of electrons = 4
so, by substitution:
㏑4.98 x 10^-5 = 4*96500*E / 8.314*298
∴E = -0.064 V
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer:
2HI + K2SO3=>2KI+H2SO3
Explanation:When aqueous hydroiodic acid and aqueous potassium sulfite are mixed the products obtained are potassium iodide and sulfurous acid.Both reactants are ionic compounds and they undergo double replacement reaction.In a double replacement reaction the parts of the ionic compounds are changed.The product is obtained by combinig cation of one compound with anion of other compound.so in above reaction sulfurous acid is obtained which is in gaseous form and potassium iodide is an ionic compound.
