22. Sodium-24 has a half-life of 15 hours. How much sodium-24 will remain in an 42.0-g sample after 90 hours?

How to do them 11-15 most difficult ones for me

BigorU [14]

Hope this can help you

6 0

3 years ago

What is the name of(NH4)3PO4

Tju [1.3M]

Answer:

What is the name of the compound (NH4) 3PO4? (NH)4+ is an ammonium radical and (PO4) 3- is a phospate radical. When we start writing the compound,the valency of phosphate goes to ammonium as subscript,and that of ammonium goes to phosphate. Hence the formula (NH4)3 PO4 and it's name is ammonium phospate.

Explanation:

5 0

3 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

If the H+ ion concentrati on in an aqueous solution is 0.0010 M, why is it not possible for the OH− ion concentration to be 1.0

spayn [35]

Answer:

im not sure sry

Explanation:

6 0

2 years ago

Circle only those of the following reactions that are redox reactions:

Citrus2011 [14]

Answer: options B,D and F

Explanation:

Since redox reactions are those which involves both oxidation and reduction

In B , Cu is oxidized and S gets reduced

D, Na gets oxidized and hydrogen gets reduced

F, carbon gets oxidized and Oxygen gets reduced

In g, there is no change in oxidation no of s in both product and Reactants is same +4

Similarly in the case of Ag and Mg.

5 0

2 years ago