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Rom4ik [11]
3 years ago
12

Combustion analysis of a 13.42-g sample of estriol (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 1

0.06 g H2O. The molar mass of estriol is 288.38 g/mol . Find the molecular formula for estriol. Express your answer as a chemical formula.
Chemistry
1 answer:
meriva3 years ago
4 0

Answer:

C18H24O3

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of estriol = 13.42g

Mass of CO2 = 36.86g

Mass of H2O = 10.06g

Molar mass of estriol = 288.38g/mol

Step 2:

Determination of the mass of Carbon (C), Hydrogen (H) and Oxygen (O) present in the compound. This is illustrated below:

For Carbon, C:

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

Mass of C in CO2 = 12/44 x 36.86 = 10.05g

For Hydrogen, H:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 10.06 = 1.12g

For Oxygen, O:

Mass of O = 13.42 – (10.05 + 1.12) = 2.25g

Step 3:

Determination of the empirical formula for estriol. This is illustrated below:

C = 10.05g

H = 1.12g

O = 2.25g

Divide by their molar mass

C = 10.05/12 = 0.8375

H = 1.12/1 = 1.12

O = 2.25/16 = 0.1406

Divide by the smallest i.e 0.1406

C = 0.8375/0.1406 = 6

H = 1.12/0.1406 = 8

O = 0.1406/0.1406 = 1

Therefore, the empirical formula for estriol is C6H8O

Step 4:

Determination of the molecular formula for estriol. This is illustrated below:

Molecular formula is simply a multiple of the empirical formula i.e

Molecular formula => [C6H8O]n

[C6H8O]n = 288.38g/mol

[(12x6) + (8x1) + 16]n = 288.38

[72 + 8 + 16]n = 288.38

96n = 288.38

Divide both side by 96

n = 288.38/96 = 3

Molecular formula => [C6H8O]n

=> [C6H8O]n

=> [C6H8O]3

=> C18H24O3

Therefore, the molecular formula for estriol is C18H24O3

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Answer:

Kelvin

Explanation:

8 0
3 years ago
1. How much heat (in calories) is needed to raise 20 g of H2O from 5°C to 40°C? (c = 1.0 cal/g °C)
KIM [24]

Answer:

700 calories

Explanation:

Using the formula below:

Q = m × c × ∆T

Where;

Q = amount of heat required (calories)

m = mass of substance (g)

c = specific heat of substance (cal/g°C)

∆T = change in temperature (°C)

According to this question, the following information was provided;

Q = ?

m = 20g

c = 1.0 cal/g °C

∆T = 40°C - 5°C = 35°C

Using the formula; Q = m × c × ∆T

Q = 20 × 1 × 35

Q = 700 calories

Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.

6 0
3 years ago
1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?
vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

  • P1= 16 atm
  • V1
  • P2= ?
  • V2= V1÷2= \frac{V1}{2} because the volume is halved.

So:

16 atm*V1= P2* \frac{V1}{2}

Solving:

\frac{16 atm*V1*2}{V1}=P2

16 atm*2= P2

32 atm= P2

<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>

2. Now

  • P1= 5.5 atm
  • V1
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  • V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

\frac{5.5 atm*V1}{3*V1}=P2

\frac{5.5 atm}{3}= P2

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Answer:

O c. One that is answered through observation

I used process of elminations so

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Only the one that scientists ask doesnt make sense

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