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Rainbow [258]
2 years ago
8

g The following equation is given for the dissociation of the complex ion Ag(NH3) 2. The dissociation constant, Kd, is the equil

ibrium constant for the dissociation of a complex ion into its component ions. What would be the expression for the dissociation constant of Ag(NH3) 2.
Chemistry
1 answer:
Sveta_85 [38]2 years ago
8 0

Answer:

Kd = [Ag⁺] × [NH₃]² / [Ag(NH₃)₂⁺]

Explanation:

Let's consider the dissociation reaction of the complex ion Ag(NH₃)₂⁺.

Ag(NH₃)₂⁺(aq) ⇄ Ag⁺(aq) + 2 NH₃(aq)

The dissociation constant, Kd, is the equilibrium constant for the dissociation of the complex ion, that is, it is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.

The dissociation constant for this reaction is:

Kd = [Ag⁺] × [NH₃]² / [Ag(NH₃)₂⁺]

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Calculate the molar mass for each of the following:
Mila [183]

Answer: a . 152g/mol  b. 102g/mol c. 183g/mol

Explanation:

By stating the atomic masses of each element in the questions, we have;

Fe= 56, S= 32, O= 16, Al = 27, C = 12, H =1 , N = 14, therefore

(a). FeSO4 =  56 + 32 + (16 x 4) =  152g/mol

(b). Al2O3 =  (27 x 2) + (16 x 3) = 102g/mol

(c). C7H5NO3S ( Saccharin, an artificial Sweetner) =

   (12 x 7) + (1 x 5) + 14 + (16 x 3) + 32 =  183g/mol

3 0
3 years ago
1pt According to Einstein's theory of gravity, the more mass an object has, the
wariber [46]

Answer:

B. more it will warp space

Explanation:

4 0
2 years ago
A block has a mass of 30g and dimensions of 1.5cm by 4.8cm by 2.3cm. Calculate the density (round to TWO sig figs).
adelina 88 [10]

Answer:

1.8g/cm³

Explanation:

Density is defined as the ratio of the mass of a substance and the space this subtance occupies. It is usually given in g/cm³.

The mass of the block is 30g.

The volume this mass occupies is 1.5cm × 4.8cm × 2.3cm = 16.56cm³.

The density is:

30g / 16.56cm³ =

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5 0
3 years ago
A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express
Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

(c) Molarity = 13792 mol HNO3/m3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\  \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3

(c) HNO3 molarity (mol HNO3/m3)

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are  1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3

We can now calculate the molarity as

Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3}  =13792 \frac{molHNO3}{m3}

8 0
3 years ago
Which element is placed in the same period as ruthenium but has a higher atomic number than it
Blababa [14]
Since it is in period five, any element that has a higher atomic number than it could be an answer, therefore, silver, bismuth, and osmium are possibilities.
8 0
3 years ago
Read 2 more answers
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