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Mekhanik [1.2K]
3 years ago
13

Why do scientists use STP? options: A. STP has no negative numbers which makes calculations easier. B. Because temperature and p

ressure affect gases, scientists want to "standardize" the conditions at which gases are studied. C.Using 500 K and 500 kPa for standard temperature and pressure, conditions are kept constant. D. STP is used so that the Celsius temperature can be used in gas law calculations
Chemistry
1 answer:
alexdok [17]3 years ago
6 0

Answer: B

Explanation: It makes the most sense to me as the way temperature is measured, is with 0 degrees Celsius as the standard and 1 atm for pressure, if I am incorrect for any reason, please let me know

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compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of
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<u>Answer:</u> The empirical and molecular formula of the compound is CH_2F and C_{14}H_{28}F_{14}  respectively

<u>Explanation:</u>

We are given:

Mass of CO_2=3.926g

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, \frac{12}{44}\times 3.926=1.071g of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100      ......(1)

  • <u>For Carbon:</u>

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%

  • <u>For Fluorine:</u>

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles

Moles of Fluorine = \frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = \frac{0.072}{2.67}=1.34\approx 1

For Hydrogen = \frac{6.36}{2.67}=2.38\approx 2

For Fluorine = \frac{2.67}{2.67}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is CH_2F

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = 12+(2\times 1)+19]=33g/mol

Putting values in above equation, we get:

n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}

Hence, the empirical and molecular formula of the compound is CH_2F and C_{14}H_{28}F_{14}  respectively

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