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3 years ago

HELP ME PLEASEEEEE!!!!!!!!!!

Chemistry

2 answers:

Marta_Voda [28]3 years ago

4 0

The answer is c because a acid when an acid is dissolved in water, the balance between hydrogen ions and hydroxide ions is shifted. Now there are more hydrogen ions than hydroxide ions in the solution. This kind of solution is acidic.

AlladinOne [14]3 years ago

3 0

Answer:

I dont know what the answer is

Explanation:

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4 years ago

The reaction of C4H8 and Br2 to yield C4H2Br2 represents

natima [27]

Answer:

A Synthesis Reaction

Explanation:

A + B --> AB

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4 years ago

What is the definition of a chemical bond?

Gnom [1K]

Answer:

Explanation:

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7 0

3 years ago

The electron configuration for chromium is 1s22s22p63s23p63d54s1 instead of 1s22s22p63s23p63d44s1. The configuration is an excep

g100num [7]

<u>Answer:</u> The correct answer is Aufbau principle.

<u>Explanation:</u>

Pauli exclusion principle states that no two electrons with in an atom can have all four quantum numbers same.

Heisenberg principle states that it is impossible to measure with high precision the value of momentum and position of an electron.

Aufbau principle states that the electron will occupy the lowest energy level first before occupying the higher energy levels.

Schrodinger equation is used to find the allowed energy levels of quantum mechanical systems of an electron.

Chromium is the 24th element of the periodic table and its electronic configuration must be written as: $1s^22s^22p^63s^23p^63d^44s^2$

But the actual configuration for this is $1s^22s^22p^63s^23p^63d^54s^1$

This configuration is an exception to Aufbau's principle because half filled sub-levels is more stable than other configurations.

The actual configuration has half filled 'd' and 's' sub-levels but the expected configuration had fully filled 's' orbital and partially filled 'd' orbital.

Thus, actual configuration is accepted for chromium atom.

Hence, the correct answer is Aufbau's principle.

7 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

4 years ago