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Pepsi [2]
2 years ago
7

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

course of their duties. One such investigator, while reorganizing their shelves, has mixed up several small vials and is unsure about the identity of a certain powder. Elemental analysis of the compound reveals that it is 67.31 % carbon, 6.978 % hydrogen, 4.617 % nitrogen, and 21.10 % oxygen by mass. Which of the compounds could the powder be?
Chemistry
1 answer:
almond37 [142]2 years ago
3 0

The powder could be acetaminophen, analgesic  having chemical formula C_{17}H_{21}NO_{4}

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule.

We are given:

Percentage of C = 67.31 %

Percentage of H = 6.978 %

Percentage of N = 4.617 %

Percentage of O = 21.10 %

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

Mass of C = 67.31 g

Mass of H = 6.978 g

Mass of N = 4.617 g

Mass of O = 21.10 g

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

Step 1: Converting the given masses into moles.

Moles of carbon = \frac{mass}{molar \;mass}

Moles of carbon =\frac{67.31g}{12g/mole}

=5.60 moles

Moles of hydrogen = \frac{mass}{molar \;mass}

Moles of hydrogen = \frac{6.978 g}{1 g/mole}

=6.978 moles

Moles of nitrogen =\frac{mass}{molar \;mass}

Moles of nitrogen = \frac{4.617  g}{14 g/mole}

=0.329 moles

Moles of oxygen =\frac{mass}{molar \;mass}

Moles of oxygen =\frac{21.10 g}{16 g/mole}

=1.31 moles

Step 2: 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 0.329 moles.

We get the ratio of C : H : N : O = 17 : 21 : 1 : 4

The empirical formula for the given compound is C_{17}H_{21}NO_{4}.

Learn more about the empirical formula here:

brainly.com/question/14044066

#SPJ1

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Carlos is making phosphorus trichloride using the equation below. He uses 15.5 g of phosphorus and collects 50.9 g of phosphorus
alex41 [277]

Answer: 35.4 g

Explanation:

The balanced reaction is :

2P+3Cl_2\rightarrow 2PCl_3

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of phosphorous}=\frac{15.5g}{31g/mol}=0.50moles

\text{Moles of phosphorous chloride}=\frac{50.9g}{137g/mol}=0.372moles

2P+3Cl_2\rightarrow 2PCl_3

According to stoichiometry :

2 moles of phosphorous chloride are produced by = 3 moles of Cl_2

Thus 0.37 moles of phosphorous chloride are produced by=\frac{3}{2}\times 0.372=0.558moles of Cl_2

Mass of Cl_2=moles\times {\text {Molar mass}}=0.558moles\times 71g/mol=35.4g

Thus 35.4 g of chlorine reacted with the phosphorus

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Potassium-40 is a radioactive isotope that decays into a single argon-40 atom and other particles with a half-life of 1:25 billi
stiks02 [169]

Answer:

0.147 billion years = 147.35 million years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of Potassium-40 is 1.25 billion years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

<em></em>

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

  • At the time needed to be determined:

<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>

  • If we start with 100% Potassium-40:

∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

  • That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.

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