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Oksi-84 [34.3K]
3 years ago
14

Click on the box to choose which compound has a greater difference in electronegativities of its members. Use the

Chemistry
2 answers:
Daniel [21]3 years ago
5 0

The two compounds are KCl or NaCl

Answer:

KCl has greater electro negativity difference.

Explanation:

Electro negativity of chlorine on the Pauling's scale is 3.16 while that of potassium is 0.82. The electro negativity difference between them is 2.34

Electro negativity of sodium is 0.93. Hence electro negativity difference in NaCl is 2.23. Hence KCl has greater electro negativity difference.

4vir4ik [10]3 years ago
3 0

Answer:

See the example below.

Explanation:

Let us make an example with a group of compounds to apply the general <em>trends of electronegativities</em> within families to find which compound has a greater difference in electronegativities.

Use this group.

  • a. HBr
  • b. HI
  • c. HF
  • d. HCl
  • e. H₂

Br, I, F, and Cl belong to the halogen family (group or column 17 of the periodic table).

Electronegativity is a relative measure of how strong an atom atracts the bonding electrons in a compound.

Since the strength of attracion of the valence electrons by the nucleus of an atom decreases as the size of the atom incease, in a family of elements (same group) the electronegativity decreases as you go down in the group.

The order of the halogens from top to bottom in the column in the periodic table is F, Cl, Br, and I.

Thus, the order of their electronegativiies is: F > Cl > Br > I. F is the most electronegative atom, and all those elements have higher electronegativy than H. (The electronegativiy of I is 2.66 and the electronegativity of H is 2.2). Thus, including H, the rank of the electronegativities is F > Cl > Br > I > H.

Thus, the greater difference in electronegativities is for HF (option a.).

Note, that the difference in electronegativites of the atoms of the molecule H₂ is zero, since both atoms have the same electronegativity.

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1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
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Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

<em>Q2:  Calculate the energy change (q) of the surroundings (water) using the enthalpy equation </em>

<em>qwater = m × c × ΔT.  </em>

<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>

<em></em>

a) First part: the energy change (q) of the surroundings (water):

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>

<em>b) second part:</em>

<em>Q water = Q unknown metal. </em>

<em>Q unknown metal =  - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

<em>Q unknown metal =  - </em>668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.

<em></em>

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