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Paha777 [63]
2 years ago
12

Write the formula for lead oxide.

Chemistry
1 answer:
serious [3.7K]2 years ago
5 0

Answer:

<h2>Lead(II) oxide</h2>

Explanation:

<h3>Lead(II) oxide, also called lead monoxide, is the inorganic compound with the molecular formula PbO. PbO occurs in two polymorphs: litharge having a tetragonal crystal structure, and massicot having an orthorhombic crystal structure. Modern applications for PbO are mostly in lead-based industrial glass and industrial ceramics, including computer components. It is an amphoteric oxide.[3]</h3>

  • IUPAC name
  • Lead(II) oxide

  • Other names
  • Lead monoxide
  • Litharge
  • Massicot
  • Plumbous oxide
  • Galena

<h2> Preparation</h2><h3>PbO may be prepared by heating lead metal in air at approximately 600 °C (1,100 °F). At this temperature it is also the end product of oxidation of other oxides of lead in air:[4]</h3><h3>Thermal decomposition of lead(II) nitrate or lead(II) carbonate also results in the formation of PbO:</h3>

<h3>2 Pb(NO</h3><h3>3)</h3><h3>2 → 2 PbO + 4 NO</h3><h3>2 + O</h3><h3>2</h3><h3>PbCO</h3><h3>3 → PbO + CO2</h3><h3>PbO is produced on a large scale as an intermediate product in refining raw lead ores into metallic lead. The usual lead ore is galena (lead(II) sulfide). At a temperature of around 1,000 °C (1,800 °F) the sulfide is converted to the oxide:[5]</h3>

<h3>2 PbS + 3 O</h3><h3>2 → 2 PbO + 2 SO2</h3><h3>Metallic lead is obtained by reducing PbO with carbon monoxide at around 1,200 °C (2,200 °F):[6]</h3>

<h3>PbO + CO → Pb + CO2</h3>

pls brainlest meh

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EL hidroxido de sodio reacciona con el sulfato de hierro (II) para formar sulfato de sdio e hidroxido ferroso. Si se hace reacci
Oduvanchick [21]

Answer:

280.8 g

Explanation:

Definimos la reaccion:

2NaOH +  FeSO₄  →  Na₂SO₄  +  Fe(OH)₂

Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.

Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol

2 moles de NaOH producen 1 mol de hidroxido ferroso

Entonces 6.25 moles producirán, la mitad (6.25  . 1) /2 = 3.125 moles

Convertimos los moles a masa:

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Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water
barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same  

The dilution formula

 M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M

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