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mojhsa [17]
3 years ago
8

use the principles of atomic structure and/or chemical bonding to explain each of the following. in each part, your answers must

include references to both substances. the atomic radius of li is larger than that of be. the second ionization energy of k is greater than the second ionization energy of ca. the carbon-to-carbon bond energy in c2h4 is greater than it is in c2h6. the boiling point of cl2 is lower than the boiling point of br2.
Chemistry
1 answer:
denpristay [2]3 years ago
4 0

The magnitude of dispersion forces in Br2 is greater than the magnitude of dispersion forces in Cl2.

Atomic radius decreases across the period but increases down the group. As more shells are added to the atom, the repulsion between electrons increases. Across the group, more electrons are added without increase in the number of shell hence atomic radius decreases across the period due to increase in the size of the nuclear charge. Therefore,  the atomic radius of Li is larger than that of Be.

Ionization energy is a periodic trend that increases across the period but decreases down the group. Since the outermost electron is further from the nucleus due to screening of inner electrons, ionization energy decreases down the group. Across the period, the size of the nuclear charge increases hence ionization energy increases across the period.

For K, the second electron is removed from an inner shell which requires a very large amount of energy. In Ca, the second electron is removed from the valence shell which requires a lesser amount of energy. Therefore, the second ionization energy of K is greater than the second ionization energy of Ca.

The carbon to carbon bond in C2H4 is a double bond which has a greater bond enthalpy than the single bond in C2H6. As such, the carbon to carbon bond in C2H4 has a greater bond energy than the carbon to carbon bond in C2H6.

The boiling point of Cl2 is lower than the boiling point of Br2 because Br2 is larger than Cl2 hence the magnitude of dispersion forces in Br2 is greater than the magnitude of dispersion forces in Cl2.

Learn more: brainly.com/question/11155928

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Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

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The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


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As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.

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5 0
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