Answer:
<em>Hello There. The answer is Both </em>provide evidence<em> for evolution. Homologous </em>structures <em>are structures that are similar in related organisms because they were inherited from a </em>common ancestor. <em>These </em>structures may or may not have the same function <em>in the descendants. </em>
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The enthalpy change of the reaction is <u>-1347.8 kJ.</u>
<h3>What is the enthalpy change, ΔH, of the reaction?</h3>
The enthalpy change, ΔH, of the reaction is calculated from Hess's law of constant heat summation as follows:
Hess's law states that the enthalpy change of a reaction is the sum of the enthalpies of the intermediate reaction.
Given the reactions below and their enthalpy values;
1. X (s) + 12 O₂ (g)⟶ XO (s) ΔH₁ = −850.5 kJ
2. XCO₃ (s) ⟶ XO (s) + CO₂ (g) ΔH₂ = +497.3 kJ
The enthalpy change, ΔH, of the reaction whose equation is given below, will be:
X (s) + 12 O₂ (g) + CO₂ (g) ⟶ XCO₃ (s)
ΔH = ΔH₁ - ΔH₂
ΔH = − 850.5 kJ - (+497.3 kJ)
ΔH = -1347.8 kJ
Learn more about enthalpy change at: brainly.com/question/14047927
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Answer:
There are 0.1 moles of solute in 250 mL of 0.4 M solution
Explanation:
because it is
<span>Generally, a hydrogen bond can be characterized as a proton shared by two lone electron pairs. It occurs when a hydrogen (H) atom, covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F), experiences the electrostatic field of another highly electronegative atom nearby.
Among the choices in the bond (-N...H-O) one side of the Hydrogen is bonded to a highly electronegative atom with a lone pair (-N) and the other side is directly bonded with a highly electronegative atom (O-).
So -N...H-O- shows a hydrogen bond.</span>
