All categories

3 years ago

Which atom has higher electron affinity F or I, Why?

1 answer:

8 0

Down a group, electron affinity typically decreases. This is because the atomic radius increases down a group.
Fluorine, which is higher up the group then chlorine, has a lower electron affinity. This is because the electrons in the outermost shell of a fluorine atom are closer together.
The electron gained also feels a great amount of repulsion from the electrons originally in the outermost shell. Energy is required to keep the gained electron in the shell, causing fluorine to have a smaller electron affinity than chlorine.
Hope this helps :)

You might be interested in

This reaction appears to be destroying the logs

Alex_Xolod [135]

What exactly is the question you are asking?

5 0

4 years ago

Use the formula for pressure to determine the weight of a box that creates a pressure of 1,065 N/cm2 when resting on a 17.0-cm2

Sloan [31]

Force /Area = Pressure
Force = Pressure * Area
Force = 1065 * 17 = 18105 N

weight = 18105 N

I hope this helps!

8 0

3 years ago

Read 2 more answers

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

<u>Answer:</u> The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. 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:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago

Please help y’all I don’t understand

DedPeter [7]

Answer:

the last one

Explanation:

6 0

3 years ago

Calculate the freezing point of a 12.25 m 12.25m aqueous solution of propanol. Freezing point constants can be found in the list

vichka [17]

Answer:

∆t(f) = 49.755C

Explanation:

Freezing point is defined as the point in which a liquid changes from liquid to solid state. Therefore, to calculate the freezing point of aqueous solution of propanol

∆t = k × I × m

Where k is cryscopic constant =

I = vant Hoff factor = 2

M = molar concentration = 12.25m

∆t (f) = 1.9 × 12.25 × 2 = 49.755C

5 0

4 years ago

Which atom has higher electron affinity F or I, Why?​

Which atom has higher electron affinity F or I, Why?