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Rudik [331]
3 years ago
9

Which statement best explains the difference between polar and nonpolar covalent bonds?(1 point)

Chemistry
2 answers:
Mars2501 [29]3 years ago
8 0

Answer:

Let me help you out fr:

Explanation:

1. 0.5 to 1.7

2. fluorine (F)

3. KNO3

4. Polar covalent bonds share electrons unequally, while nonpolar covalent bonds share electrons equally.

5. Fe and S have an ionic bond, while S and O have covalent bonds.

miv72 [106K]3 years ago
6 0

Answer:

Answer is 4

Explanation:

Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared between two atoms. ... If the electronegativity of two atoms is basically the same, a nonpolar covalent bond will form, and if the electronegativity is slightly different, a polar covalent bond will form.

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melting point will be higher than that of pure ethyl acetate

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Based on the shapes of the water drops in part D, how does the attractive force between water and wax compare to the attractive
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4 0
2 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
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