Answer:
5. Atoms with high ionization energies and high electron affinities have low electronegativities.
Explanation:
Ionization energy is the minimum amount of energy which is required to knock out the loosely bound valence electron from the isolated gaseous atom.
Electron affinity is the amount of energy released when an isolated gaseous atom accepts electron to form the corresponding anion.
Electronegativity is the tendency of an atom in a bond pair to attract the shared pair of electron towards itself.
Low ionization energies as well as low electron affinities mean the atom has low effective nuclear charge, which results in the less attraction of the valence electrons by the atom and thus, low electronegativity.
Answer:
m = 0.531 molal
Explanation:
∴ m fructose = 3.35 g
∴ V water = 35.0 mL
∴ ρ H2O = 1 g/mL
- molality = moles solute / Kg solvent
∴ Mw fructose = 180.16 g/mol
⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose
⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O
⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O
⇒ m = 0.531 molal
Hello!
answer : c
eg = octahedral, mg = square planar, sp3d2
Hope that helps!
Answer:
43.75 g of Nitrogen
Explanation:
We'll begin by calculating the mass of 1 mole of NH₄NO₃. This can be obtained as follow:
Mole of NH₄NO₃ = 1 mole
Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)
= 14 + 4 + 14 + 48
= 80 g/mol
Mass of NH₄NO₃ =?
Mass = mole × molar mass
Mass of NH₄NO₃ = 1 × 80 = 80 g
Next, we shall determine the mass of N in 1 mole of NH₄NO₃.
Mass of N in NH₄NO₃ = 2N
= 2 × 14
= 28 g
Thus,
80 g of NH₄NO₃ contains 28 g of N.
Finally, we shall determine the mass of N in 125 g of NH₄NO₃. This can be obtained as follow:
80 g of NH₄NO₃ contains 28 g of N.
Therefore, 125 g of NH₄NO₃ will contain = (125 × 28) / 80 = 43.75 g of N.
Thus, 125 g of NH₄NO₃ contains 43.75 g of Nitrogen
