Diagram A shows the Lewis structure (LS) of CH_2O. The formal charge on each atom is zero.
To get the formal charge (FC) on the atoms, cut each bond in half, as in <em>Diagram B</em>. Each atom gets the electrons on its side of the cut.
Formal charge = valence electrons in isolated atom - electrons on bonded atom
FC = VE - BE
<em>On O:
</em>
VE = 6
BE = 2 lone pairs 2 + 2 bonding electrons = 4 + 2 = 6
FC = 6 – 6 = 0.
<em>On H:
</em>
VE = 1
BE = 1 bonding electron
FC = 1 – 1 = 0
<em>On C:
</em>
VE = 4
BE = 1 in each single bond + 2 in the double bond = 2 + 2 = 4
FC = 4 - 4 = 0
Answer:
The answers to your question are below
Explanation:
a) 6.85×1020 H2O2 molecules
H2O2 MW = 32 + 2 = 34 g
34g -------------------- 6.023 x 10²³ molecules
x ------------------- 6.85 x 10 ²⁰
x = (6.85 x 10 ²⁰)(34)/ 6.023 x 10²³
x = 0.038 g
3.3×1022 SO2 molecules
MW SO2 = 32 + 32 = 64g
64 g -------------------- 6.023 x 10²³ molecules
x -------------------- 3.3×1022 SO2 molecules
x = (3.3×1022 SO2)(64) / 6.023 x 10²³
x = 3.51 g
5.5×1025 O3 molecules
MW = 16 x 3 = 48g
48 g ----------------- 6.023 x 10²³ molecules
x ------------------ 5.5×1025 O3 molecules
x = (5.5×1025 )(48) / 6.023 x 10²³
x = 4383 g
9.30×1019 CH4 molecules
MW = 12 + 4 = 16 g
16 g -------------------- 6.023 x 10²³ molecules
x -------------------- 9.30×1019 CH4 molecules
x = (9.30×1019)(16) / 6.023 x 10²³
x = 0.0025 g
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
Answer:
5200 ppm
Explanation:
As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.
Given in the question,
Water = 250 ml = 250 g
Lead = 1.30 g
So,
ppm of Lead = 
= 
= 5200 ppm
So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.
6.02214086 x 10^23 mol^-1
