hooc are carboxyl groups
your r or amino groups are those unique structures which have different atoms in them. your nh2 groups are your hydrogen atoms
21.69mL - 20.70mL = .99mL Fe
7.8 g / .99 mL = 7.9g/mL
Answer:
a. Polar
b. Polar
c. Non-polar
d. Non-polar
Explanation:
a.
, hydronium cation contains a positive charge. Just as any other ion, it is polar, as it has a net charge.
b.
has the same shape as water. There are two lone pairs on sulfur atom which produce an overall dipole moment in this molecule, the bent structure is polar.
c.
is non-polar, as the central atom, phosphorus, doesn't contain any lone pairs, all the dipole moments cancel out: two dipole moments in the vertical plane, P-Cl, and three P-Cl dipoles in the horizontal plane within a trigonal bipyramidal shape.
d.
is non-polar, since it's a tetrahedral molecule with no lone pairs on carbon atom, all four C-F dipole moments cancel out to yield a net 0 dipole moment.
Answer:
The melting point of the solid is higher in temperature then it is compared to the freezing point of the liquid.
Explanation:
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>