Answer:
sorry don't know the answer but i really need the points sorry
Explanation:
I believe that the answer is 1.8^24 of Ni atoms in 3.6 mol of Ni.
Hope this helps. :)
Atomic number is 8 and atomic mass is taken as 16 amu
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
mass of HNO₃ = 0.378 g
Explanation:
Normality = Molarity * number of equivalents
Molarity = Normality/number of equivalents
normality of HNO₃ = 0.30 N, Volume = 20 mL
HNO₃ ionizes in the following way:
HNO₃(aq) ----> H⁺ + NO₃⁻
Therefore, number of equivalents for HNO₃ is 1
molarity of HNO₃ = 0.30/1 =0.30 mol/dm³
Using the formula, molarity = number of moles/volume in liters
number of moles = molarity * volume
Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL
number of moles = 0.006 moles
From the formula, mass = number of moles * molar mass
molar mass of HNO₃ = 63.0 g/mol
mass = 0.006 * 63
mass of HNO₃ = 0.378 g
