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Setler79 [48]
3 years ago
5

The electronegativity of C = 2.5, O = 3.5. What type of bond is between C-O? ______ ______ bond

Chemistry
1 answer:
Schach [20]3 years ago
5 0

Answer:

Polar covalent bond

Explanation:

On the basis of electronegativity bond could be ionic bond, polar and non pole covalent bond.

Ionic bond:

It is the bond which is formed by the transfer of electron from one atom to the atom of another element.  

Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.  The electronegativity difference between bonded atoms is greater than 1.7.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Both atoms are bonded together electrostatic attraction occur between anion and cations.

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.  

The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.  There are two type of covalent bond. Polar and non polar covalent bond. When electronegativity difference is 0.4 - 1.7 bond is polar covalent bond when it is less than 0.4 the bond is non polar covalent.

For example:

In CO the electronegativity of oxygen is 3.5 and carbon is 2.5. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and carbon becomes partial positive. and bond is polar covalent.

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The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C
Dmitry [639]

Answer:

-1160kj/mol

Explanation:

the reaction is exothermic because heat is released to the environment

7 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
jelani and Mikah are watching an approaching storm from their living room window. All of a sudden they see a bright flash of lig
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4 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
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