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
D = m / V
d = 1300 g / 743 cm³
d = 1.749 g/cm³
The third option, 2,2,1,2
Answer:
The atoms of some elements share electrons because this gives them a full valence shell just like the nobal gases GOOD LESSONS ♡
<span>0.0687 m
The balanced equation is
BaCl2 + Na2SO4 ==> BaSO4 + 2 NaCl
Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights.
Atomic weight sodium = 22.989769
Atomic weight sulfur = 32.065
Atomic weight oxygen = 15.999
Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol
Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol
Molarity is defined as moles per liter, so let's do the division.
0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m
Rounding to 3 significant figures gives 0.0687 m</span>
