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luda_lava [24]
3 years ago
15

The antibiotic doxycycline is used to treat Lyme disease. A 100 mg dose of the antibiotic consists of 59.5 mg of C, 5.40 mg of H

, 6.30 mg of N and 28.8 mg of O by mass. What is the empirical formula of doxycycline
Chemistry
1 answer:
never [62]3 years ago
5 0

Answer:

C₁₁H₁₂NO₄

Explanation:

In order to determine the empirical formula of doxycycline, we need to follow a series of steps.

Step 1: Determine the centesimal composition

C: 59.5 mg/100 mg × 100% = 59.5%

H: 5.40 mg/100 mg × 100% = 5.40%

N: 6.30 mg/100 mg × 100% = 6.30%

O: 28.8 mg/100 mg × 100% = 28.8%

Step 2: Divide each percentage by the atomic mass of the element

C: 59.5 /12.0 = 4.96

H: 5.40/1.00 = 5.40

N: 6.30/14.0 = 0.450

O: 28.8/16.0 = 1.80

Step 3: Divide all the numbers by the smallest one

C: 4.96/0.450 = 11

H: 5.40/0.450 = 12

N: 0.450/0.450 = 1

O: 1.80/0.450 = 4

The empirical formula of doxycycline is C₁₁H₁₂NO₄

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Which of the following would most likely result in an increase in reaction rate? placing the reactants on a hotplate
puteri [66]

Answer:

placing the reactants on a hot plate

Explanation:

If the temperature goes up, the reaction rate will increase. Because the particle will move faster and makes the kinetic energy larger.

4 0
3 years ago
what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan
PSYCHO15rus [73]
<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

  1. Mass of Lithium as 1.50 g
  2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

  • First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

  • Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

                   = 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

                                     = 0.054 moles

  • According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
  • Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
  • On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.

7 0
3 years ago
The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mariulka [41]

Answer:

3.09kg

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 from the balanced equation = 2 x 114 = 228g

Converting 228g of C8H18 to kg, we obtained:

228/1000 = 0.228kg

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Converting 704g of CO2 to kg, we obtained:

704/1000 = 0.704kg

From the equation,

0.228kg of C8H18 produced 0.704kg of CO2.

Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2

6 0
3 years ago
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