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PIT_PIT [208]
3 years ago
5

On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential

energy of two O atoms versus the distance between their nuclei.

Chemistry
1 answer:
ki77a [65]3 years ago
6 0

Answer:

Explanation:

We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.

From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.

Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.

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The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is
WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

  • sp^{3}, four electron domains, as in ethane \text{C}_2\text{H}_6
  • sp^{2}, three electron domains, as in ethene \text{C}_2\text{H}_4
  • sp, two electron domains, as in ethyne \text{C}_2\text{H}_2

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

  • Carbon atoms with a sp^{3} hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately 109.5\textdegree{}
  • Carbon atoms with a sp^{2} hybridization would demonstrate a triangular planar configuration with a bond angle of 120\textdegree{}
  • Carbon atoms with a sp hybridization would demonstrate a linear configuration with a bond angle of 180\textdegree{}

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

  • sp^{3} in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
  • sp^{2} in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.
3 0
4 years ago
I need to know the answers to this to make sure I am right ​
oee [108]
The answer is C............
3 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
3 years ago
Has earth always had the same land masses and oceans that it has today? What will happen in the future, will they stay in the sa
icang [17]

Answer: The look of the Earth has changed over time but continents have always been part of the picture. They didn't always look the way they do today, but yes, there have always been continents on Earth. Earth's continents are constantly in motion, and by running the tape ... against southern Europe, while the Atlantic will be a far wider ocean than it is today. ... During the war, Hess had used sonar to map some areas of the ocean floor in detail. ... away from both sides of the mid-ocean ridge at the same rate.

7 0
3 years ago
Read 2 more answers
Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.
Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

2C(gr) +2H_{2} (g) +O_{2} (g)  → CH_{3} COOH(l)

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

C(gr) +H_{2} (g) +O_{2} (g) → CH_{3} COOH(l)

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

2C(gr) +2H_{2} (g) +O_{2} (g)  → CH_{3} COOH(l)

The combustion of liquid acetic acid is given as,

CH_{3} COOH(l) + 2O(g) → 2CO_{2}((g) +2H_{2} O(l)                    ΔH =-873

learn more about  balancing chemical equation

brainly.com/question/15052184

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5 0
2 years ago
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