Which metal could you use to reduce Cr3+ ions but not Mn2+ ions?

inna [77]

Answer:

Explanation:

The two green substances are not the same. Most of their properties are different, except some are alike. Since not all of their properties are matching, they are not the same substance.

I am not sure if this is what you were looking for, but if it is then I am happy to help.

7 0

2 years ago

when 0.72 g of a liquid is vaporized at 110° C and 0.967 atm, the gas occupies a volume of 0.559L. The empirical formula of the

Leno4ka [110]

Hydrogen gas mixed with sulfur

8 0

3 years ago

In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equat

artcher [175]

Answer:do you still need help ?

Explanation:

5 0

2 years ago

Read 2 more answers

Which of the following statements are true regarding the dissociation of Al(NO3)3? Select more than one

Gelneren [198K]

Answer:

There will be one Al3+ ion.

There will be 3 NO3- ions

Explanation:

Dissociation equation:

Al(NO₃)₃ → Al³⁺ + 3NO₃¹⁻

When aluminium nitrate dissociate it produces one silver ion (Al³⁺) and three (NO₃¹⁻) ions.

Properties of Al(NO₃)₃:

It is inorganic compound having molecular mass 169.87 g/mol.

It is white odor less compound.

Its density is 4.35 g/mL.

Its melting and boiling points are 120°C and 440°C.

It is soluble in water.

It is sued to treat infections.

It is used in the photographic films.

It s toxic and must be handled with great care.

4 0

2 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

3 years ago