Cs and Cl would likely form an ionic compound true or false?

Chemistry

1 answer:

kipiarov [429]3 years ago

6 0

True. Cs is metal, Cl is nonmetal, the difference in electronegativities is high enough to form ionic bond.

You might be interested in

Thermoelectric Bridge CER: Design and Draw

miss Akunina [59]

Answer:

roar

Explanation:

UUUUUUUUUUWWWWWWWWWWWUUUUUUUUU

8 0

2 years ago

3 reasons why aluminium is used in making of cooking vessels

Annette [7]

Answer is : It is a good thermal and electrical conductor. -The main point to be noted is that aluminium is a highly reactive element and still it is used for making cooking utensils. The reason is that aluminium has a very high affinity for oxygen. So, it reacts with oxygen and forms a layer of aluminium oxide on its surface.

7 0

2 years ago

Read 2 more answers

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$