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kakasveta [241]
4 years ago
15

You have 1.0 mole of each compound below. which has the greatest mass?

Chemistry
1 answer:
kramer4 years ago
5 0
(a) Iron (iii) sulphate:
From the periodic table:
mass of iron = 55.845 grams
mass of sulphur = 32.065 grams
mass of oxygen = 16 grams
Iron (iii) sulphate has the formula: Fe2(SO4)3
molar mass = 2(55.845) + 3(32.065) + 3(4)(16) = 399.885 grams

(b) Sodium hydroxide:
From the periodic table:
mass of sodium = 22.989 grams
mass of oxygen = 16 grams
mass of hydrogen = 1 gram
Sodium hydroxide has the formula: NaOH
molar mass = 22.989 + 16 + 1 = 39.989 grams

(c) Barium carbonate
From the periodic table:
mass of barium = 137.327 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Barium carbonate has the formula: BaCO3
molar mass = 137.327 + 12 + 3(16) = 197.327 grams

(d) ammonium nitrate:
From the periodic table:
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
Ammonium nitrate has the formula: NH4NO3
molar mass = 14 + 4(1) + 14 + 3(16) = 80 grams

(e) Lead (iv) oxide
From the periodic table:
mass of lead = 207.2 grams
mass of oxygen = 16 grams
Lead (iv) oxide has the formula: PbO2
molar mass = 207.2 + 2(16) = 239.2 grams

From the above calculations, we can see that:
Iron (iii) sulphate has the greatest mass.
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Germanium has a face-centered cubic unit cell. The density of germanium is 5.32 g/cm3. Calculate a value for the atomic radius o
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Answer:

1.59x10⁻¹⁰m

Explanation:

To solve this question we must know that the length of the cubic cell, X, is equal to:

X = √8 * R

<em>Where R is the atomic radius of germanium</em>

And that in 1 unit cell there are 4 atoms of germanium.

To solve this question we must find the mass in 1 unit cell, with this mass we can find the volume of the cube and the length. With the length we can know the atomic radius:

<em>Mass in 1 unit cell -Molar mass Ge = 72.64g/mol:</em>

4 atoms Ge * (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles Ge

6.64x10⁻²⁴ moles Ge * (72.64g / mol) = 4.825x10⁻²²g Ge

<em>Volume unit cell:</em>

4.825x10⁻²²g Ge * (1cm³ / 5.32g) = 9.07x10⁻²³cm³

<em>Length unit cell:</em>

∛9.07x10⁻²³cm³ = 4.49x10⁻⁸cm * (1m / 100cm) = 4.49x10⁻¹⁰m

<em>Atomic radius Ge:</em>

4.49x10⁻¹⁰m / √8 =

<h3>1.59x10⁻¹⁰m</h3>
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