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emmainna [20.7K]

3 years ago

9

A solution of which of the following coordination compounds will form a precipitate when treated with aqueous AgNO3?

1 answer:

lidiya [134]3 years ago

3 0

Answer:

[Cr(NH3)6.]C13

Explanation:

Alfred Werner's coordination theory (1893) recognized two kinds of valency;

Primary valency which are nondirectional and secondary valency which are directional.

Hence, the number of counter ions precipitated from a complex depends on the primary valency of the central metal ion in the complex.

We must note that it is only these counter ions that occur outside the coordination sphere that can be precipitated by AgNO3.

If we consider the options carefully, only [Cr(NH3)6.]C13 possess counter ions outside the coordination sphere which can be precipitated when treated with aqueous AgNO3.

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Nitrogen gas is more abundant and I miss her than oxygen however nicer than to be converted into different forms to be used by m

docker41 [41]

Answer:

who lives in Barclays what?? this makes no sense

Explanation

8 0

2 years ago

A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from

tia_tia [17]

Answer:

The correct answer is: X is nitrogen dioxide, and Y is a metal oxide

Explanation:

Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.

The general equation is given as:

$4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x$

Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.

6 0

3 years ago

Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take

tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = 165°C

Depression in freezing point = $\Delta T_f=?$

$\Delta T_f=T- T_f=179^oC-165^oC=14^oC$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( organic compounds)

$\Delta T_f=14^oC$

$14^oC=1\times 40^oC kg/mol\times m$

$m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg$

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0

3 years ago

PLZ HELP!<br><br><br>How does changing the angle of a ramp affect the force required?

galina1969 [7]

Answer:

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Explanation:

7 0

3 years ago

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0

3 years ago