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3 years ago

What family are the elements carbon, germanium and lead in?

Chemistry

2 answers:

snow_tiger [21]3 years ago

7 0

Answer: Group 14 is the carbon family/group

grin007 [14]3 years ago

5 0

Answer:

Damian here!! :))

The carbon family consists of the elements carbon (C), silicon (Si), germanium (Ge), tin (Sn), lead (Pb), and flerovium (Fl). Atoms of elements in this group have four valence electrons. The carbon family is also known as the carbon group, group 14, or the tetrels. Elements in this family are of key importance for semiconductor technology.

Explanation:

Hope this helps? :))

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a. silicon b. carbon c. beryllium d.chromium

zepelin [54]

The answer is silicon because it’s atomic number is 14

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Michelle learns in science class that simple machines such as an inclined plane can change the amount of force needed to lift he

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Answer: the amount of force on the spring scale

Explanation:

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2 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

3 years ago

Bath salts are typically composed of the ingredients listed below. Identify each item as being acidic, basic, or neutral when di

Gwar [14]

Answer:

NaCl neutral

Na₂SO₄ basic

NaHCO₃ basic

K₃PO₄ basic

Na₃C₆H₅O₇ basic

CaCl₂ neutral

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water.

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water

Explanation:

Identify each item as being acidic, basic, or neutral when dissolved in water. If a particular ingredient does make an acidic or basic solution, describe how this occurs.

NaCl

When NaCl is dissolved in water the resulting solution is neutral.

Na₂SO₄

When Na₂SO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the SO₄²⁻ ion.

SO₄²⁻(aq) + H₂O(l) ⇄ HSO₄⁻(aq) + OH⁻(aq)

NaHCO₃

When NaHCO₃ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the HCO₃⁻ ion.

HCO₃⁻(aq) + H₂O(l) ⇄ H₂CO₃(aq) + OH⁻(aq)

K₃PO₄

When K₃PO₄ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the PO₄³⁻ ion.

PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

Na₃C₆H₅O₇ (sodium citrate)

When Na₃C₆H₅O₇ is dissolved in water the resulting solution is basic due to the basic hydrolysis of the C₆H₅O₇³⁻ ion.

C₆H₅O₇³⁻(aq) + H₂O(l) ⇄ C₆H₆O₇²⁻(aq) + OH⁻(aq)

CaCl₂

When CaCl₂ is dissolved in water the resulting solution is neutral.

Sodium oxalate (Na₂C₂O₄) dissolves and dissociates in water according to the following equation:

Na₂C₂O₄(aq) ⇄ 2 Na⁺(aq) + C₂O₄²⁻(aq)

Oxalic acid (C₂H₂O₄) dissolves and dissociates in water according to the following equation:

C₂H₂O₄(aq) ⇄ C₂HO₄⁻(aq) + H⁺(aq)

4 0

2 years ago

Why does celsium have a lower first ionzation energy than rubidiumn.

densk [106]

Answer:

i really have no clue

Explanation:

8 0

2 years ago