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2 years ago

What is the most abundant element on Earth?

Chemistry

1 answer:

Liula [17]2 years ago

6 0

Answer:

oxygen is the most abundant element on the earth.

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6. dissolving ice tea mix in water

Cloud [144]

Answer:

Dissolving ice tea mix in water is a chemical change

Explanation:

It is a chemical change because you can not bring back the ice tea mix back if you mix it in water.

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3 years ago

Briefly explain formality in solution.

Zielflug [23.3K]

Answer:

Both molarity and formality express concentration as moles of solute per liter of solution. Formality is a substance's total concentration in solution without regard to its specific chemical form. ... The formality of a solution is defined as the number of formula mass of any solute dissolved in 1 litre of solution.

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3 years ago

Which of the following occurs when an electron moves from a higher energy shell to a lower energy shell?

AlexFokin [52]

When an electron moves from a higher energy shell to a lower one, energy in the form of light, more specifically a wavelength of a photon is released.

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3 years ago

Binomial nomenclature example and sentence using that word<br><br> please help im going to cry

Alexandra [31]

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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:

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

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