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djyliett [7]
3 years ago
14

An unknown element, X, has an atomic mass of 107.868 amu. The X-109 isotope (108.905 amu) is 48.16%. What is the amu of the othe

r isotope (report final answer to the correct number of significant figures)
Chemistry
1 answer:
juin [17]3 years ago
4 0

Answer:

106.905 amu is the mass of the other isotope

Explanation:

The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:

X = X-109*i + X-107*i

Where X is the atomic mass = 107.868 amu

X-109 = 108.905amu, i = 48.16% = 0.4816

X-107 = ?, i = 1-0.4816 = 0.5184

Replacing:

107.868amu = 108.905amu*0.4816 + X-107*0.5184

55.4194 = X-107*0.5184

106.905 = X-107

<h3>106.905 amu is the mass of the other isotope</h3>
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A 7.028 gram sample of a sodium sulfate hydrate is heated. After driving off all the water, 3.100 grams of the anhydrous salt is
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Answer:

NaSO_4.10H_2O

Explanation:

Given that:-

Mass of the hydrated salt = 7.028 g

Mass of the anhydrous salt = 3.100 g

Mass of water eliminated = Mass of the hydrated salt - Mass of the anhydrous salt = 7.028 - 3.100 g = 3.928 g

<u>Moles of water: </u>

Mass of water = 3.928 g

Molar mass of H_2O = 18 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus, moles are:

moles= \frac{3.928\ g}{18\ g/mol}

moles_{water}= 0.2212\ mol

<u>Moles of anhydrous salt: </u>

Amount = 3.100 g

Molar mass of NaSO_4 = 142.04 g/mol

Thus, moles are:

moles= \frac{3.100\ g}{142.04\ g/mol}

moles_{CaSO_4}= 0.02182\ mol

The simplest ratio of the two are:

NaSO_4:H_2O =0.02182 :  0.2212 = 1 : 10

<u>Hence, the formula for hydrate is:- NaSO_4.10H_2O</u>

7 0
3 years ago
What is the balanced equation for the combustion of sulfur
Alex17521 [72]

Answer:

There are three possible chemical equations for the combustion of sulfur:

  • 2S (s)  + O₂ (g)  → 2SO (g)

  • S (s) + O₂ (g) → SO₂ (g)

  • 2S (s) + 3O₂ (g) → 2SO₃ (g)

Explanation:

<em>Combustion</em> is a reaction with oxygen. The products of the reaction are oxides, and energy is released in the form of heat and light.

<em>Sulfur</em> iis a nonmetal, so the oxide formed is a nonmetal oxide.

The most common oxidation numbers of sulfur are -2, + 2, + 4, and + 6.

The combination of sulfur with oxygen may be only with the positive oxidation numbers (+2, + 4, and +6).

Then you have three different equations for sulfur combustion:

<u>1) Oxidation number +2:</u>

  • S(s) + O₂(g) → SO(g)

Which when balanced is: 2S(g) + O₂(g) → 2SO(g)

<u>2) Oxitation number +4:</u>

  • S(s) + O₂(g) → SO₂(g)

That equation is already balanced.

<u>3) Oxidation number +6:</u>

  • S(s) + O₂(g) → SO₃(g)

Which when balanced is: 2S(s) + 3O₂(g) → 2SO₃(g)

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4 years ago
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