Define the term allotropy and name the two allotropes of sulphur.

Chemistry
Define the term allotropy and name the two allotropes of sulphur.

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Answer

18.4\mol\dm−318.4 \mol\dm^{-3}

a)
Allotropy is the existence of two or more different physical forms (allotropes) of the same element in the same physical state. These forms differ in physical properties but have similar chemical properties.

b)
The two allotropes of sulphur are:
rhombic sulphur and monoclinic sulphur.

9
Concentrated sulphuric(VI) acid (\ceH2SO4\ce{H2SO4}) is 98%98\% by mass and has a density of 1.841.84 g cm−3^{-3}. Calculate its molarity.

Molar mass of \ceH2SO4=2(1)+32+4(16)=98\ce{H2SO4} = 2(1) + 32 + 4(16) = 98 g mol−1^{-1}.

Step 1: Mass of 11 dm3^3 (10001000 cm3^3) of solution.

Mass=density×volume=1.84 gcm−3×1000 cm3=1840 g.Mass = density \times volume = 1.84 \, g cm^{-3} \times 1000 \, cm^3 = 1840 \, g.

Step 2: Mass of \ceH2SO4\ce{H2SO4} in 11 dm3^3 of solution.

Massof\ceH2SO4=98%×1840 g=0.98×1840=1803.2 g.Mass of \ce{H2SO4} = 98\% \times 1840 \, g = 0.98 \times 1840 = 1803.2 \, g.

Step 3: Moles of \ceH2SO4\ce{H2SO4}.

Moles=1803.2 g98 gmol−1=18.4 mol.Moles = \frac{1803.2 \, g}{98 \, g mol^{-1}} = 18.4 \, mol.

Step 4: Molarity.

Molarity=18.4 mol1 dm3=∗∗18.4 moldm−3∗∗.Molarity = \frac{18.4 \, mol}{1 \, dm^3} = **18.4 \, mol dm^{-3}**.

18.4 \mol\dm^{-3}

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a) Allotropy is the existence of two or more different physical forms (allotropes) of the same element in the same physical state.

Define the term allotropy and name the two allotropes of sulphur.
Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

ScanSolve AI Answer
a) Allotropy is the existence of two or more different physical forms (allotropes) of the same element in the same physical state. These forms differ in physical properties but have similar chemical properties. b) The two allotropes of sulphur are: rhombic sulphur and monoclinic sulphur. 9 Concentrated sulphuric(VI) acid (H2SO4) is 98\% by mass and has a density of 1.84 g cm^-3. Calculate its molarity. Molar mass of H2SO4 = 2(1) + 32 + 4(16) = 98 g mol^-1. Step 1: Mass of 1 dm^3 (1000 cm^3) of solution. Mass = density × volume = 1.84 \, g cm^-3 × 1000 \, cm^3 = 1840 \, g. Step 2: Mass of H2SO4 in 1 dm^3 of solution. Mass of H2SO4 = 98\% × 1840 \, g = 0.98 × 1840 = 1803.2 \, g. Step 3: Moles of H2SO4. Moles = 1803.2 \, g98 \, g mol^-1 = 18.4 \, mol. Step 4: Molarity. Molarity = 18.4 \, mol1 \, dm^3 = 18.4 \, mol dm^-3. 18.4 ^-3