What name should be used for the ionic compound
LiI?
The name that should be used for
the ionic compound LiI is Lithium iodide. When naming ionic compounds, you have
to take into consideration pairing of two components.
Answer:
The position of an equilibrium always shifts in such a direction as to relieve a stress applied to the system -Le Chåtelier's principle
A molecule that donates a proton when it encounters a proton acceptor.- Bronsted-Lowry acid
Occurs when a reaction involving an acid and its conjugate base is combined with a second reaction involving a base and its conjugate acid.- Neutralization
It ionizes completely when dissolved in water.- Strong acid
The shift in the position of equilibrium caused by the addition of a participating ion.- Common ion effect
It only partially ionizes when dissolved in water.- Weak electrolyte
It is capable of acting as either an acid or a base depending upon the solute- Amphoteric solvent
The act of self-ionization of a solvent to produce both a conjugate acid and a conjugate base.- Auto-protolysis
A chemical species that bears both positive and negative charges.- Zwitterion
Explanation:
In the answer box we have various chemical terminologies and their definitions. In answering the question, you must carefully read through each definition, then check what option best matches that definition from the options provided.
Each definition applies only to one terminology as you can see in the answer above.
Answer:
Explanation:
01: 5.3316+6.87+37.48
02: 12.2016+37.48
03: 49.6816
04: 49.68
Answer: 49.68 (Decimals: 2; Significant Figures: 4)
HOPE THIS HELPS..
- The reaction in which we can't take back the product to reactant form .Or in simple words the reaction which is irreversible is called chemical reaction
Example:-
We cannot pull back the reaction to get wax back .
Answer: 116 g of copper
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr = 
(1hr=3600s)
of electricity deposits 63.5 g of copper.
352800 C of electricity deposits = 
of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles
