Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,
When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation, 
= 102 pm
Radius of cation, 
= 196 pm
As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
It matches the universal pH indicator and is indicating the proper pH
Explanation:
to find POH u need PH first
so ph=-log(h3o)
so
Answer:
b. Na+ leaves the salt bridge and enters enters the cathode
Explanation:
A galvanic cell or electrochemical cell depicts an oxidation -reduction half reactions (redox) reaction. it consists of two half cells ; one for the reduction reaction which involves the gain of electrons and the other for the oxidation reaction which involves the loss of electrons. One half cell contains the anode and oxidation occurs at the anode while the other half cell contains the cathode and reduction occurs at the cathode. The anode is usually connected to the cathode, a salt bridge is added to complete the circuit and allow current to flow. The salt bridge serves as a counter ions, they do not interfere with the electrochemical reaction but provides a passage for the migration of ions thereby preventing the cells from reaching equilibrium too quickly and thus the electrons in the salt are able to move along with any electrons.
In this galvanic cell, Cu at the anode losses two electrons to become Cu2+, and the electrons moves from the anode to the cathode where Mg2+ gain these two electrons to become negatively charged. Positively charged ions in the salt brigde Na+ will move to the cathode to pick negatively charged ions from the cathode solution. this helps to remove the strong negative charge from the cathode and allows the electrons to continue to move to the cathode.
