Each Be–Cl bond is polar because the two atoms have different electronegativities. The number of outer atoms (2) and lone pairs on the central atom (0) indicate that this molecule has a linear geometry. The bonds in a linear molecule are symmetric, and so their dipoles cancel out.
Each O–H bond is polar because the two atoms have different electronegativities. The number of outer atoms (2) and lone pairs on the central atom (2) indicate that this molecule has a bent geometry. The bonds in a bent molecule are asymmetric, and so their dipoles do not cancel out. In addition, the asymmetric arrangement of the lone pairs on O further contribute to the dipole of this molecule.
An O–O or O=O bond is nonpolar because the two atoms have the same electronegativity. Because there is no overall polarity in O2, the molecule is nonpolar.
Explanation:
is used for making ropes, used for climbing rocks and for making parachutes. Their usage shows that nylon fibres have high tensile strength
Answer:
9.25
Explanation:
Let first find the moles of 
and 
number of moles of = 0.40 mol/L × 200 × 10⁻³L
= 0.08 mole
number of moles of = 0.80 mol/L × 50 × 10⁻³L
= 0.04 mole
The equation for the reaction is expressed as:
The ICE Table is shown below as follows:
Initial (M) 0.08 0.04 0
Change (M) - 0.04 -0.04 + 0.04
Equilibrium (M) 0.04 0 0.04
for buffer solutions
since they are in the same solution
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
