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iris [78.8K]
3 years ago
7

1) Draw a particle picture of liquid lemonade turning into a solid popsicle BEFORE AFTER

Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Explanation:

The particles are further apart in liquids than in solids so when the lemonade changes from a liquid to a solid, the particles would become closer.

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Please explain:) I would really appreciate a step by step explanation if possible.
UNO [17]

The enthalpy change of the reaction, ΔH = -311 kJ

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

<h3>What is the enthalpy change for the reduction of ethyne to form ethane?</h3>

The enthalpy change for the reaction is obtained from the summation of the enthalpies of the reactions of the intermediate steps according to Hess's law.

The equation of the reaction is given below:

  • C₂H₂ + 2 H₂ → C₂H₆

The enthalpy of the reaction, ΔH = ΔH₁ + 2ΔH₂ + (-ΔH₃)

ΔH = {(-1299) + (2 * -286) + (1560)}Kj

ΔH = -311 kJ

The equation for the methanation reaction is given below:

3 H₂O + CO → CH₄ + H₂O

The enthalpy for the methanation reaction is as follows:

ΔH = 1.5ΔH₁ + 0.5*(-ΔH₂) + ΔH₃ + -ΔH₄

ΔH = (-483.6 * 1.5) + (0.5 * 221.0) + (-802.7) + (393.5)

ΔH = -1024.1 kJ/mol

Molar mass of CO = 28 /mol

Enthalpy change involved in the reaction of 300 g of CO = 300/28 * -1024.1 kJ/mol

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

In conclusion, the enthalpy changes are calculated from the enthalpy values of the  intermediate reactions.

Learn more about enthalpy changes at: brainly.com/question/26991394

#SPJ1

7 0
2 years ago
One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 × 105 Pa undergoes a reversible adiabatic compressi
oksian1 [2.3K]

Answer:

  • final temperature (T2) = 748.66 K
  • ΔU = w = 5620.26 J
  • ΔH = 9367.047 J
  • q = 0

Explanation:

ideal gas:

  • PV = RTn

reversible adiabatic compression:

  • δU = δq + δw = CvδT

∴ q = 0

∴ w = - PδV

⇒ δU = δw

⇒ CvδT = - PδV

ideal gas:

⇒ PδV + VδP = RδT

⇒ PδV = RδT - VδP = - CvδT

⇒ RδT - RTn/PδP = - CvδT

⇒ (R + Cv,m)∫δT/T = R∫δP/P

⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303

∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5

⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212

⇒ T2/T1 = 2.512

∴ T1 = 298 K

⇒ T2 = (298 K)×(2.512)

⇒ T2 = 748.66 K

⇒ ΔU = Cv,mΔT

⇒ ΔU = (3/2)R(748.66 - 298)

∴ R = 8.314 J/K.mol

⇒ ΔU = 5620.26 J

⇒ w = 5620.26 J

  • H = U + nRT

⇒ ΔH = ΔU + nRΔT

⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)

⇒ ΔH = 5620.26 J + 3746.787 J

⇒ ΔH = 9367.047 J

8 0
3 years ago
Which is a form of heat transfer that is reduced by using a potholder when taking a hot dish out of an oven? Please answer quick
Angelina_Jolie [31]

Answer:

i belive your talking about Conduction

Explanation:

if  im

wrong comment

3 0
2 years ago
What is the molar Mass of 2.4g of oxygen
liq [111]
<span>2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g of CO2 and 1.44g of H2O. The relative molecular mass of the compound was found to be 60. a)What are the masses of carbon, hydrogen and oxygen in 2.4g of the compound? b)What are the emperical and ..</span>
4 0
2 years ago
When forming a chemical bond only the ________ electrons participate in the bonding process.
matrenka [14]
The outer shell electrons are only involved in the bonding process since they are the only 'incomplete' shell and it needs to be fulfilled by another element. 
8 0
3 years ago
Read 2 more answers
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