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2 years ago

HELP PLEASE ASAP

Chemistry

1 answer:

Anna [14]2 years ago

3 0

Answer:

Explanation:

Aluminium and Indian are in the same group/family

Sodium and Sulfer are in the same period

Elements of the same family have similar properties

Since sulfur and sodium aren't in the same family they dont share properties

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Molecular compounds are capable of forming multiple covalent bonds. the compound below that contains a triple bond is:

alexira [117]

The answer is <u>CO(g).</u>

4 0

1 year 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

3 years ago

Why does galium melt

levacccp [35]

Answer:

Like most other metals, Gallium is solid at room temperature (or liquid if it is too hot in your room). But, if it is held [in hands] for long enough, it melts in your hands, and doesn't poison you like Mercury would. This is because of its unusually low melting point of (~29 degree Centigrade).

- It melts once it reaches its melting point.

7 0

3 years ago

These values also increase from the bottom to the top of a group because the size of the atom decreases, resulting in a distance

Allushta [10]

<span>These values also increase from the bottom to the top of a group because the size of the atom decreases, resulting in a smaller distance between the nucleus and the valence electron shell, which increases the attraction between the protons and electrons.</span>

7 0

3 years ago

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

guajiro [1.7K]

Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)

Kp = p SO₃² / ( p SO₂² x p O₂ )

The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm SO₂ O₂ SO₃

I 3.3 0.79 0

C -2x -x 2x

E 3.3 - 2x 0.79 - x 2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:

p SO₂ = 3.3 -0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm = .56 atm

Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

7 0

3 years ago