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Gala2k [10]
2 years ago
14

the combustion of 17.41 grams of a compound known to only contain carbon, hydrogen, and oxygen produced 25.61 grams of carbon di

oxide and 10.48 grams of water. a.) how much percentage of the compound is carbon? b.) how much percentage of the compound is hydrogen? c.) how much percentage of the compound is oxygen?
Chemistry
1 answer:
Svet_ta [14]2 years ago
3 0

The compound which on combustion produces Carbon-di-oxide and water is carbohydrates.

  • Percentage of Carbon = 6.98 / 17.41 x 100 = 40 %
  • Percentage of Hydrogen = 15.08 / 17.41 x 100 = 86.61 %
  • Percentage of Oxygen = 4.6 / 17.41 x 100 = 26.42 %

<h3>What is combustion reaction ?</h3>

Those reaction in which the Crabon present in the reatctant gets reacted with oxygen and produces Carbon-di-oxide.


It is an Exothermic Reaction.

<h3>Combustion of carbohydrate ;

C_6H_1_2 O_6 + O_2    - > 6CO_2 +2H_2O</h3>

To find out the percentage of C, H and O, we must know about there weight in the product.

Given,

Total weight of compund [Reactant] = 17.41 grams

Weight of carbon dioxide [Product] = 25.61 grams

Weight of water [Product]= 10.48 grams

Now, we need to find out the weight of C,H and O in the products ;

  • Mass of Carbon = No. of Moles x Atomic weight


No. of Moles of Crabon =   \frac{Given \\ weight\\  of\\  the\\  compound}{Molecular\\ weight\\  of\\  the compound} x No. of C-atoms in molecule                          

No. of Moles of Crabon =25.61 / 44 = 0.58 x 1 = 0.58 moles

Mass of Carbon = 0.58 moles x 12 = 6.98 gm

  • Mass of Hydrogen = No. of Moles x Atomic weight

No. of Moles of Hydrogen =   \frac{Given \\ weight\\  of\\  the\\  compound}{Molecular\\ weight\\  of\\  the compound} x No. of H-atoms in molecule                          

No. of Moles of Hydrogen =10.48/ 18 = 0.58 x 2 = 1.16 moles

Mass of Hydrogen = 1.16 moles x 12 = 15.08 gm

  • Mass of Oxygen = Total weight of compound - (Weight of C & H)

                                       =   17.41 grams - ( 15.08 gm + 6.98 gm)
Mass of Oxygen  = 4.6 gm

Now,

Percentage of atom =  \frac{Mass \\ of \\ Atom}{Total\\  weight \\ of\\  compound}   x 100

  • Percentage of Carbon = 6.98 / 17.41 x 100 = 40 %
  • Percentage of Hydrogen = 15.08 / 17.41 x 100 = 86.61 %
  • Percentage of Oxygen = 4.6 / 17.41 x 100 = 26.42 %

Learn more about Percentage composition here ;

brainly.com/question/20065048

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One problem with elimination reactions is that mixtures of products are often formed. For example, treatment of 2-bromo-2-methyl
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Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

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7 0
2 years ago
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