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vagabundo [1.1K]
2 years ago
8

write the names of the following compounds. 1) iron + sulphur . 2) manganese + oxygen. aluminium + sulphur​

Chemistry
2 answers:
Vlada [557]2 years ago
8 0

Answer:

  1. FeS
  2. MnO₂
  3. Al₂S₃

Explanation:

<u>1) Iron + Sulphur = ?</u>

→ Iron sulfide

<u>2) Manganese + Oxygen = ?</u>

→ Manganese dioxide

<u>3) Aluminium + Sulphur = ?</u>

→ Aluminium sulfide

Tatiana [17]2 years ago
3 0

#1

  • Ferrous shows variable valency as it's transitive metal.

#For valency=2

\\ \rm\Rrightarrow {Fe(II)\atop Iron}+{S\atop Sulphur}\longrightarrow {FeS\atop Pyrrhotite}

#For valency=3

\\ \rm\Rrightarrow {Fe(III)\atop Iron}+{S\atop Sulphur}\longrightarrow {Fe_2O_3\atop Sesquisulfide}

#2

  • Manganese also shows variable valency

#For valency=2

\\ \rm\Rrightarrow {Mn(II)\atop Manganese}+{O_2\atop Oxygen}\longrightarrow {MnO_2\atop Manganese (II) Oxide}

#For valency=4

\\ \rm\Rrightarrow {Mn(IV)\atop Manganese}+{O_2\atop Oxygen}\longrightarrow {MnO_4\atop Permanganate}

#3

\\ \rm\Rrightarrow {Al\atop Aluminium}+{S\atop Sulphur}\longrightarrow {Al_2S_3\atop Aluminium\:Sulphide}

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Explanation:

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
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The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

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The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

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Ca(OH)2+ CO2------ CaCO3

when excess of carbon dioxide is passed through calcium carbonate calcium hydrogen carbonate is formed and solution become colourless.

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