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Fofino [41]
3 years ago
11

What are the formal charges on the sulfur (s), carbon (c), and nitrogen (n) atoms, respectively, in the resonance structure that

contributes most to the stability of the thiocyanate ion, scn−?

Chemistry
2 answers:
sergiy2304 [10]3 years ago
7 0

Answer:

Sulfur: -1

Carbon: 0

Nitrogen: 0

Explanation:

The thiocyanate ion SCN- can have only two resonance structures, which are:

S - C ≡ N <--------> S = C = N

In the first structure, we have one single bond and one triple bond, in this case, the negative charge is located in the sulfur. This is because Sulfur have 6 electrons and those electrons are present in the atom, (see picture below), and counting the electron that is sharing with the Carbon, the total electrons that sulfur has is 7 (It has one more than usual). Carbon and nitrogen are already stable with 0 of formal charge, because carbon can only have 4 electrons which 1 is sharing with sulfur and the other 3 with the nitrogen, and nitrogen have 5 electrons, three sharing with carbon and the other two kept it for itself.

In the second structure, the negative charge of the sulfur is transfered to the nitrogen, meaning that it has 6 electrons the nitrogen (formal charge -1) and carbon and sulfur with 4 and 6 electrons respectively.

Between these two structures, the most stable is the first one basically because Sulfur is a better nucleophile than the Nitrogen, and can form stronger hydrogen bond in acid, giving more stable structure.

mrs_skeptik [129]3 years ago
3 0

Answer:

0, 0, -1  

Explanation:

There are three resonance contributors to the resonance hybrid of the SCN⁻ ion.

The one on the right is the least stable because it has the greatest separation of charge.

The remaining structures each have one negative charge.

The structure in the middle is the most stable and most important contributor because it has the negative charge on the most electronegative atom —  N.

The formal charges on S, C, and N are 0, 0, and -1.

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14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km
emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0
3 years ago
Why do you think it is advantageous to use inert electrodes in the electrolysis process? ​
olga nikolaevna [1]

Answer:

The advantage of this technique is that purified water as well as deposited metals can be re-used. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode.

5 0
2 years ago
Another name for the acid test ratio is the
elena55 [62]
Liquidity metric is the other name
4 0
3 years ago
calculate q for the following: 125.0 ml of 0.0500 m pb(no3)2 is mixed with 75.0 ml of 0.0200 m nacl at 25oc chegg
alexandr402 [8]

The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.

Given, T = 25°C.

<h3>Chemical equation:</h3>

Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2

PbCl2 in aqueous solution split into following ions

PbCl2 ------ Pb(+2) + 2Cl-

Q = [Pb(+2)] [Cl-]^2

The Concentration of Pb(+2) ions and Cl- ions can be calculated as

[Pb(+2)] = 0.06 × 125/200

= 0.0375

[Cl-] = 0.02 × 75/200

= 0.0075

By substituting all the values, we get

[0.0375] [0.0075]^2

= 2.11 × 10^(-6).

Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

learn more about Ions:

brainly.com/question/13692734

#SPJ4

6 0
2 years ago
What is the empirical formula for the following molecular formula: C10H5O2
Tju [1.3M]

The empirical formula is the same as the molecular formula : C₁₀H₅O₂

<h3>Further explanation</h3>

Given

Molecular formula : C₁₀H₅O₂

Required

The empirical formula

Solution

The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.  

The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

<em>(EF)n=MF </em>

(EF)n = C₁₀H₅O₂

If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula

4 0
3 years ago
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