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:
See explanation
Explanation:
1 mole of a gas occupies 22.4 L
x moles occupies 16.8 L
x = 1 mole * 16.8 L/22.4 L
x = 0.75 moles
number of moles = mass/molar mass
mass = number of moles * molar mass
mass = 0.75 moles * 30.01 g/mol = 22.5075 g = 2.25 * 10^1 g
the coefficient of the scientific notation answer = 2.25
the exponent of the scientific notation answer = 1
significant figures are there in the answer = 6
the right most significant figure in the answer = 3
2.
number of moles = 12.5g/38g/mol = 0.3289 moles
1 mole occupies 22.4 L
0.3289 moles occupies 0.3289 moles * 22.4 L/1 mole
= 7.36736 L = 7.36736 * 10^0 L= 7.37 * 10^0 L
the coefficient of the scientific notation answer =7.37
the exponent of the scientific notation answer = 0
significant figures are there in the answer = 6
the right most significant figure in the answer= 3
Answer:
D. Electron pairs repelling each other push atoms apart
Explanation:
Hope this helps :) I just got it wrong on ap3x so I'm sure this is right
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The answer to this is solved through stochiometry: the answer is this: 0.0833mol
