Omg yes her creepiness makes her 20x better
Hey there!
A ) PO₄³⁻
The element exhibits resonance (4 times) as there is a double bondwith 1 oxygen and a single with the other 3 oxygen. This switchesin each resonance structure. (I know its weird b/c I leaves Phosphorus with 10 electrons surrounding it)
b) SO₄²⁻
Sulfur is the central atom ,double bond 2 oxygens so that they can only have 4 valence electrons ( the little dots around the oxygen ) ,single bond 2 oxygens ( So that they have 6 valence electrons )
Formal charges are 0 for the two double bonded oxygens [ 6-6 = 0 ]and 0 for the sulfur [ 6-6=0 ]
But the oxygens are -1 [ 6-7 = -1 ]
Since there are two of them, there is a negative charge of -2
Bracket the whole structure and on the top right of the outside ofthe brackets write 2- and include a - (negative sign) next to eachof the single bonded oxygens
Don't forget there are resonance structures. The double bonds canbe placed around any of the oxygens
Hope that helps!
For the answer to the question above,
in this problem we can use the concept of freezing point depression since we are given the freezing point of the solution.
ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
<span>m = 0.4014
</span>So the answer to your question is m = 0.4014
A. 0.0105 kg
b. 1570 m
c. 3.5x10^-6
d. 3500000
e. 1000 mL
f. 0.000358 m3
g. 548.6 cm3
Answer:
Rate law for the uncatalyzed reaction:
Rate = k[Ce^4+]^2[Tl]
Explanation: rate law shows the mathematical relationship by comparing the rate of reaction with the reactants concentration.
Rate law= k[reactantA][reactantB] .
Rate law can be used to derive an equation that shows a reactant as a function of time. Also, it can be use to calculate the rate of reaction from known concentration.