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3 years ago

How many sulfur atoms are present in 100 grams of this compound? Report your answer to three significant figures.

Chemistry

1 answer:

kap26 [50]3 years ago

7 0

Answer:

1.88 × 10²⁴ atoms

Explanation:

Step 1: Given data

Mass of sulfur: 100 g

Step 2: Calculate the moles corresponding to 100 g of sulfur

The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:

100 g × (1 mol/32.07 g) = 3.12 mol

Step 3: Calculate the number of atoms in 3.12 moles of sulfur

We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.

3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms

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Give two examples of an ionic compound. Chemical name and chemical structure.

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NaCl (Sodium chloride)
LiF (Lithium fluoride)

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Rusting is an example

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Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according

Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

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2 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2

Nezavi [6.7K]

Answer:

The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

1) Chemical reaction.

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

2) Mole ratio

2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

3) Starting amount of oxalic acid

mass = 28 mg = 0.028 g
molar mass = 90.03 g/mol
Convert mass in grams to number of moles, n:

n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol

4) Titration

Volume of base: 28.4 mL = 0.0248 liter
Concentration of base: x (unknwon)

Number of moles of acid: 2.52 mol (calculated above)

Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

$\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622$

That means that there are 0.000622 moles of NaOH (solute)

5) Molarity of NaOH solution

M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using three signficant figures, such as the starting data are reported.

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