Tutor profile: Alexia J.

What are the formal charges on the nitrogen and oxygen in the following structure? H3C-C(triple bonded to)N-O(with 3 sets of lone pairs)

The equation you will be using is the one for formal charge: formal charge=(valence e-)-(e- in lone pair +1/2 the number of bonding e-) Let's start with determining the formal charge for N Nitrogen has 5 valence e-'s, 0 lone pairs, (1/2*8 e-'s) 5-(+4)=+1 formal charge Now let's do the same thing with O: Oxygen has 6 valence e-'s, 6 lone pairs, (1/2*2 e-'s) 6-(6+1)=-1 formal charge

What volume of 0.540 M NaOH solution contains 15.5 g of NaOH?

First we need to convert g of NaOH into moles of NaOH. To do that, we need to get the molecular weight of NaOH. You find the molecular weight by adding up all individual elements' atomic mass to get a molecular weight of 40.0 g. The math will look something like this: 15.5 g NaOH x (1 mol/40 g NaOH)= .3875 mol NaOH Then we will use the molarity equation to find the volume (M=mol/L) However, since we are looking for volume, we will rearrange the equation to solve for volume (L=mol/M) Then we will plug in the numbers and solve L=.3875 mol/.540 M L=.718 L

What is the oxidation number of the sulfur atom in H2SO4?

H atoms typically have a +1 charge. Since there are 2 H atoms, we multiply the +1 charge by the 2 atoms to get a +2 charge. Oxygen atoms typically have a -2 charge. Since there are 4 O atoms, the -2 charge gets multiplied by the 4 atoms to get us -8. Since H2S04 is a neutral molecule, we know that the overall oxidation number is 0. If we set up an equation where the total oxidation number equals 0, then we can find out what the oxidation number of sulfur is. (+2)+(x)+(-8)=0. When we solve, we find out that x, or the oxidation number of sulfur is +6.