Answer:
number 7 or 3,5
Explanation:
sana po makatulong po sa inyo
Missing question:
Chemical reaction: H₂ <span>+ 2ICl → 2HCl + I</span>₂.
t₁ = 5 s.
t₂ = 15 s.
c₁ = 1,11 M = 1,11 mol/L.
c₂ = 1,83 mol/L.
rate of formation = Δc ÷ Δt.
rate of formation = (c₂ - c₁) ÷ (t₂ - t₁).
rate of formation = (1,83 mol/L - 1,11 mol/L) ÷ (15 s - 5 s).
rate of formation = 0,72 mol/L ÷ 10 s.
rate of formation = 0,072 mol/L·s.
Answer: Option (a) is the correct answer.
Explanation:
A miscible solution is defined as the one in which two or more than two components are soluble with each other.
So, when hexane is added to heptane then both of them will mix with each other and hence they are miscible. This mixing will lead to formation of more number of ions into the solution as a result, more will be the disorder present in the solution.
As entropy is the degree of randomness present.
Hence, we can conclude that a solution of hexane and heptane will lead to an increase in entropy.
The molecules or atoms that are formed by gain or loss of one or more valence electrons are said to be ions.
When atom loss one or more valence electrons, results in formation of cation whereas when atom gain one or more valence electrons, then formation of anion occurs. Cations carry positive charge and anions carry negative charge.
In general, cations are smaller than the neutral atoms from which they are formed and anions are larger than the neutral atoms.
As cations are smaller than the related neutral atoms because the valence electrons are lost which are farthest away from the nucleus. After that, taking more electrons distant from the cation results in reduction of radius of the ion.
Thus, aluminium cation consist of few electrons which results in fewer occupied energy levels by the electrons further results in reduction of radius i.e. smaller size.
Hence, given statement is true i.e. aluminium atom is larger than the aluminium cation as cation has fewer occupied energy levels.
Get to know first how many moles in the gas:n = pV/RT= (1.013*10^5*750/760) Pa *1.49*10^-3 m^3/(8.314 J/(molK)*298) n = 0.0601 moles.
The combustion energies are 889 kJ/mol (methane) and 2 220 kJ (propane) x = moles methane, y = moles propane
x*889 + y*2220 = 778 x + y = 0.0601----------- x = 0.267784 moles = 0.267784*100/0.0601 = 44.6 % y = 0.243216 moles = 0.243216*100/0.0601 = 55.4 %
