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2 years ago

How many different sets of equivalent protons are there for para-xylene (1,4-dimethylbenzene)? only 1 two three four?

Chemistry

1 answer:

ELEN [110]2 years ago

7 0

Answer:

Two

Step-by-step explanation:

There are two ways to determine the number sets of equivalent protons in a compound.

(a) By substitution

Replace each H with a different group and see if you get a different compound

For example, the six CH₃ H atoms are an equivalent set, because, if I replace any one of them, I will always get the same compound: 1-chloromethyl-4-methylbenzene (Fig. 1).

Similarly, the four aromatic H atoms are an equivalent set. If I replace any one of them, I will always get the same compound: 2-chloro-1,4-dimethylbenzene (Fig. 2).

Thus, there are two sets of equivalent protons in p-xylene.

(b) By symmetry

Two atoms are equivalent if one can be converted into the other by a symmetry operation on the molecule.

The methyl hydrogens can be interconverted by a combination of reflections about the AB and CD mirror planes and by rotations about the C-C bonds to the ring (Fig. 3).

Similarly, the four aromatic H atoms can be interconverted by a combination of reflections about the AB and CD mirror planes.

Again, we find two sets of equivalent protons.

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Please Balanced this Equation

Scilla [17]

Answer:

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$ .

Explanation:

Consider the oxidation state on each of the element:

Left-hand side:

O: -2 (as in most compounds);
Cr: $\displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6$ ;
Fe: +2 (from the charge of the ion);

Right-hand side:

Cr: +3;
Fe: +3.

Change in oxidation state:

Each Cr atom: decreases by 3 (reduction).
Each Fe atom: increases by 1 (oxidation).

Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.

Assume that the coefficient of the most complex species $\rm Cr_2O_7^{2-}$ is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.

Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain

two Cr atoms,
six Fe atoms, and
seven O atoms.

O atoms seldom appear among the products in acidic environments; they rapidly combine with $\rm H^{+}$ ions to produce water $\rm H_2O$ . Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen $\rm H^{+}$ ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$ .

7 0

2 years ago

Given the following: [G3P] = 1.5x10-5M; [BPG] = 3.0x10-3M ; [NAD+] = 1.2x10-5M; [NADH]=1.0x10-4 ; [HPO42-]= 1.2x10-5 M; pH = 7.5

mel-nik [20]

Answer: The given reaction is non-spontaneous in nature.

Explanation:

To calculate the $H^+$ concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

pH of the solution = 7.5

$7.5=-\log [H^+]$

$[H^+]=10^{-7.5)=3.1\times 10^{-8}M$

For the given chemical equation:

$\text{Glyceraldehyde3-phosphate }+NAD^++HPO_4^{2-}\rightarrow \text{1,3-Biphosphoglycerate }+NADH+H^+$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 6.3 kJ/mol = 6300 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = Temperature = $25^oC=[273+25]K=298K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{[BPG][NaDH][H^+]}{[G_3P][NAD^+][HPO_4^{2-}]}$

$[BPG]=3.0\times 10^{-3}M$

$[NADH]=1.0\times 10^{-4}M$

$[H^+]=3.1\times 10^{-8}M$

$[G_3P]=1.5\times 10^{-5}M$

$[NAD^+]=1.2\times 10^{-5}M$

$[HPO_4^{2-}]=1.2\times 10^{-5}M$

Putting values in above equation, we get:

$\Delta G=6300J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(3.0\times 10^{-3})\times (1.0\times 10^{-4})\times (3.1\times 10^{-8})}{(1.5\times 10^{-5})\times (1.2\times 10^{-5})\times (1.2\times 10^{-5})}))\\\\\Delta G=9917.02J/mol=9.92kJ/mol$