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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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Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
How many grams of NaCl are required to make 150.0 mL of a 5.000 m solution
Oksi-84 [34.3K]

The  grams   of NaCl  that are required  to  make  150.0 ml of  a  5.000 M  solution is  43.875 g


calculation

Step 1:calculate  the  number of moles

moles =  molarity  x volume  in L

volume  = 150 ml / 1000 = 0.15 L

= 0.15 L  x 5.000  M  = 0.75  moles

Step 2:  calculate mass

mass =  moles x  molar mass

molar mass  of NaCl = 23 + 35.5 = 58.5 mol /L


mass is therefore =0.75  moles  x 58.5  mol /l =43.875 g

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