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Alecsey [184]
2 years ago
4

Peroxodisulfate ion, s2o82−, oxidizes iron(ii) ion to iron(iii) ion in aqueous solution and is itself reduced to sulfate ion. wr

ite a balanced equation.
Chemistry
1 answer:
DIA [1.3K]2 years ago
8 0
The balanced equation for the reaction is;
2Fe2+(aq) + S2O8^2-(aq)  = 2SO4^2-(aq) + 2Fe3+(aq)
 This reaction occurs during the reaction of peroxodisulfate ions and iodide ions in aqueous form to form sulfate ions and iodine, where iron ii ions or iron iii ions acts as a catalyst.
Iron II ions is oxidized to iron III ions while peroxodisulfate(VI) is reduced to sulfate (VI) ions.
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How are exothermic and endothermic reactions linked in the process of refining metal ore?
Alex777 [14]
The intended sense is that of a reaction that depends on absorbing heat if it is to proceed. The opposite of an endothermic process is an exothermic process, one that releases "gives out" energy in the form of heat
6 0
2 years ago
Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui
Furkat [3]

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

\frac {8.63}{100}\times x=28.02

solving for x, we get:

<u>x = 324.18 g/mol</u>

6 0
2 years ago
Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)
Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0
2 years ago
Elements that do not have full outer electron shells will donate, share, or take electrons from other atoms. Choose the items th
Tamiku [17]

it´s actually Lithium and fluorine / Magnesium and Chlorine / Beryllium and Nitrogen

8 0
2 years ago
For a ternary solution at constant T and P, the composition dependence of molar property M is given by: M = x1M1 + x2M2 + x3M3 +
AveGali [126]

Answer:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i}) ...1

M^{\alpha } = M_{i} + CX_{xjxk}          ...2

Explanation:

The ternary constant is given by the following equation:

The symbol XiXi, where XX is an extensive property of a homogeneous mixture and the subscript ii identifies a constituent species of the mixture, denotes the partial molar quantity of species ii defined by

M_{i}  = [\frac{d(nM)}{dn_{i} }]_{P,t,n,j}

This is the rate at which property  X  changes with the amount of species  i  added to the mixture as the temperature, the pressure, and the amounts of all other species are kept constant.  A partial molar quantity is an intensive state function.  Its value depends on the temperature, pressure, and composition of the mixture.

In a multi phase system (in this case, a ternary system), the components resolved give:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i})

and M^{\alpha } = M_{i} + CX_{xjxk}

5 0
2 years ago
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