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3 years ago

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

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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>

Set up: $\displaystyle 30 \ g \ SiO_2(\frac{1 \ mol \ SiO_2}{60.09 \ g \ SiO_2})$
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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Explanation:

In this, we can use De Broglie’s equation. This equation is the relationship between De Broglie’s wavelength, velocity and the mass of a moving object. In this equation, we are using plank's constant which is 6.626 x 10^-34 m^2 kg/s.

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3 years ago

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2 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

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Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

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3 years ago