The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
![\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5B2%2A%286%2A413%20%2B%20347%29%20%2B%207%2A498%20-%20%284%2A2%2A799%20%2B%206%2A2%2A467%29%5D%20kJ%2Fmol%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
Oxygen and Carbon dioxide
Explanation:
Oxygen is required for respiration whereby energy is released from natural occurring nutrients accompanied by the release of water and carbon dioxide. carbon dioxideis also required by plants to photosynthesise.
Oxygen and carbon dioxide in the troposphere supports life as by enabling respiration in organisms and photosynthesise in plants can. Without oxygen in an environment, only life forms that live by anaerobic respiration will thrive. This affects a regions carrying capacity
Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice 
= latent heat of ice
Latent heat of ice = 334 KJ
⇒ = 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.
Zn, Cd, and Ag are transition metals that usually form only one monoatomic cation.
A monatomic cation is a cation made of only one atom.
Cations are positively charged ions, in this example Ag⁺, Cd²⁺ and Zn²⁺.
These cations form only one type of ion, while iron and copper form more than one type of cations.
Iron and copper form cations with different charges (Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).
It depends on electron configuration which type would be formed.
Electron configuration of zinc atom: ₃₀Zn 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
Transition metals are elements in the d-block of the Periodic table.
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<h3>Answer:</h3>
Strontium (Sr)
<h3>Explanation:</h3>
The condition given in statement is the presence of two valence electron. Hence, first we found the electronic configuration of given atoms as follow;
Rubidium [Kr] 5s¹
Strontium [Kr] 5s²
Zirconium [Kr] 4d² 5s²
Silver [Kr] 4d¹⁰ 5s¹
From above configurations it is cleared that only Strontium and Zirconium has two electrons in its valence shell.
We also know that s-block elements are more reactive than transition elements due to less shielding effect in transition elements hence, making it difficult for transition metals to loose electrons as compared to s-block elements. Therefore, we can conclude that Strontium present in s-block with two valence electrons is the correct answer.
