РЬ(он)2<br> +<br> Hсі —<br> Н2О +<br> РЬСl2

Write the concentrated sulphuric acid H2C2O42H2O=2H2O+CO2+CO

masha68 [24]

Answer:

H2C2O4.2H20 → CO2 + CO + H2O

Explanation:

Oxalic acid crystals are nothing but dehydrated oxalic acid (H2C2O4 . 2H2O).

On heating, the water of crystallization is lost first. Then, the dehydrated oxalic acid decomposes into carbon dioxide(CO2), carbon monoxide(CO) and water(H2O).

Equations involved :

H2C2O4 . 2H2O → H2C2O4 + 2H2O

H2C2O4 → CO2 + HCOOH (FORMIC ACID)

HCOOH → CO + H2O

Overall equation : H2C2O4.2H20 → CO2 + CO + H2O

4 0

3 years ago

when copper sulfide is partially roasted in air (reaction with o2), copper sulfite is formed first. subsequently, upon heating,

Lyrx [107]

The question is incomplete, complete question is:

When copper(I) sulfide is partially roasted in air (reaction with oxygen), copper(I) sulfite is formed first. subsequently, upon heating, the copper sulfite thermally decomposes to copper(I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.

Answer:

The balanced chemical equations for these two reactions:

$2Cu_2S(s)+3O_2(g)\rightarrow 2Cu_2SO_3(s)$

$Cu_2SO_3(s)\overset{heat}\rightarrow Cu_2O(s)+SO_2(g)$

Explanation:

On partial roasting of copper sulfide in an air. The balanced chemical reaction is given as:

$2Cu_2S(s)+3O_2(g)\rightarrow 2Cu_2SO_3(s)$

On further heating of copper(I) sulfite it get decomposes into copper oxide and sulfur dioxide. The balanced chemical reaction is given as:

$Cu_2SO_3(s)\overset{heat}\rightarrow Cu_2O(s)+SO_2(g)$

5 0

2 years ago

Explain how you can use an atoms mass number and atoic number to determine the number of protrons electrons and neutrons in the

VMariaS [17]

Simple!

Atomic number= number of protons AND number of electrons (in an atom)

Mass number= number of neutrons

7 0

3 years ago

How does the electron arrangement change moving vertically down a group?

vampirchik [111]

Answer:

Electrons arrangement change is almost constant down the group, hence nuclear charge is almost constant.

But the energy levels on which electrons are arranged increase down the group, hence shielding or screening effect increases.

6 0

2 years ago

Calcule el volumen de disolución de LiOH a 3.5 M, necesario para neutralizar una disolución de 25 ml de H2SO3 cuya densidad es d

Olin [163]

Respuesta:

0.11 L

Explicación:

Paso 1: Escribir la ecuación balanceada

2 LiOH + H₂SO₃ ⇒ Li₂SO₃ + H₂O

Paso 2: Calcular la masa de solución de H₂SO₃

25 mL de solución de H₂SO₃ tiene una denisdad de 1.03 g/mL.

25 mL × 1.03 g/mL = 28 g

Paso 3: Calcular la masa de H₂SO₃ en 26 g de Solución de H₂SO₃

La riqueza de H₂SO₃ es 60%, es decir, cada 100 g de solución hay 60 g de H₂SO₃.

26 g Sol × 60 g H₂SO₃/100 g Sol = 16 g H₂SO₃

Paso 4: Calcular los moles correspondientes a 16 g de H₂SO₃

La masa molar de H₂SO₃ es 82.07 g/mol.

16 g × 1 mol/82.07 g = 0.19 mol

Paso 5: Calcular los moles de LiOH que reaccionan con 0.19 moles de H₂SO₃

La relación molar de LiOH a H₂SO₃ es 2:1. Los moles de LiOH que reaccionan son 2/1 × 0.19 mol = 0.38 mol.

Paso 6: Calcular el volumen de solución de LiOH

0.38 moles de LiOH están en una solución 3.5 M. El volumen requerido es:

0.38 mol × 1 L/3.5 mol = 0.11 L

7 0

2 years ago

РЬ(он)2 + Hсі — Н2О + РЬСl2