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LuckyWell [14K]
2 years ago
12

HEEEEELP ASAPP IM AWARDING 30 points!!!!

Chemistry
2 answers:
alexira [117]2 years ago
5 0

Answer:

\rm S^{2-}.

Explanation:

Based on the electron configuration of this ion, count the number of electrons in this ion in total:

2 + (2 + 6) + (2 + 6) = 18.

Each electron has a charge of (-1).

Atoms are neutral and have 0 charge. However, when an atom gains one extra electron, it becomes an ion with a charge of (-1). Likewise, when that ion gains another electron, the charge on this ion would become (-2).

The ion in this question has a charge of (-2). In other words, this ion is formed after its corresponding atom gains two extra electrons. This ion has 18 electrons in total. Therefore, the atom would have initially contained 18 - 2 = 16 electrons. The atomic number of this atom would be 16.

Refer to a modern copy of the periodic table. The element with an atomic number of 16 is sulphur with atomic symbol \rm S. To denote the ion, place the charge written backwards ("2-" for a charge of (-2)) as the superscript of the atomic symbol:

\rm S^{2-}.

attashe74 [19]2 years ago
4 0

Answer:

100% Silicon

Explanation:

Hope this helps!!

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If i’m correct it’s b, bouyance force.
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Write the formula for lead oxide.
serious [3.7K]

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

5 0
1 year ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

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Contrast mixtures with pure substances
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  • have a definite and constant composition
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