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Aleks [24]
3 years ago
14

How many moles of sand (SiO2) are in 30 g of sand?

Chemistry
2 answers:
luda_lava [24]3 years ago
6 0

Answer:

\boxed {\boxed {\sf About \ 0.5 \ moles \ of \ SiO_2}}

Explanation:

To convert from grams to moles, we must use the molar mass.

<u>1. Molar Mass </u>

Use the Periodic Table to find the masses of the individual elements (silicon and oxygen) in sand.

  • Silicon (Si): 28.085 g/mol
  • Oxygen (O): 15.999 g/mol

Examine the formula for sand: SiO₂. There is a subscript of 2 after oxygen, so there must be 2 oxygen atoms. Multiply oxygen's mass by 2 and add silicon's mass to find the molar mass of sand.

  • SiO₂: 2(15.999 g/mol) + 28.085 g/mol= g/mol

<u>2. Calculate Moles </u>

Use the molar mass as ratio.

\frac{60.083 \ g \ SiO_2}{1 \ mol \ SiO_2}

Multiply by the given number of grams (30)

30 \ g \ SiO_2 *   \frac{60.083 \ g \ SiO_2}{1 \ mol \ SiO_2}

Flip the fraction so the grams of sand will cancel.

30 \ g \ SiO_2 *\frac{1 \ mol \ SiO_2}{60.083 \ g \ SiO_2}

30  *\frac{1 \ mol \ SiO_2}{60.083 }

\frac{30 \ mol \ SiO_2}{60.083 }

0.499309289 \ mol \ SiO_2

<u>3. Round </u>

The original measurement of grams has 1 signfiicant figure. We must round our answer to 1 sig fig.

For the answer we found, that is the tenth place. The 9 in the hundredth tells us to round the 4 to a 5.

\approx 0.5 \ mol \ SiO_2

There are about <u>0.5 moles</u> of SiO₂ in 30 grams.

anastassius [24]3 years ago
5 0
<h3>Answer:</h3>

0.50 mol SiO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

30 g SiO₂ (sand)

<u>Step 2: Identify Conversions</u>

Molar Mass of Si - 28.09 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of SiO₂ - 28.09 + 2(16.00) = 60.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 30 \ g \ SiO_2(\frac{1 \ mol \ SiO_2}{60.09 \ g \ SiO_2})
  2. Multiply/Divide:                  \displaystyle 0.499251 \ mol \ SiO_2

<u>Step 4: Check</u>

<em>Follow sig figs and round. We are given 2 sig figs.</em>

0.499251 mol SiO₂ ≈ 0.50 mol SiO₂

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3 years ago
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Measuring with a ruler and using final volume minus initial volume

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volume=length*width*height

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8 0
3 years ago
When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio
arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>

<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>

<em>m is molality of the solution</em>

<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>

<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>

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