The question is improperly formatted.
What is the concentration of H+ ions in a 2.2 M solution of HNO3.
Answer:-
2.2 moles of H+ per litre
Explanation:-
M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.
Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.
(3) loses one electron and becomes positively charged
Lithium has one valence electron and Bromine has seven. Therefore Lithium will give up its one to Bromine for both to have an octet
Entropy change is defined only along the path of an internally reversible process path.
<h3><u>What is Entropy Change </u>?</h3>
- Entropy is a measure of a thermodynamic system's overall level of disorder or non-uniformity. The thermal energy that a system was unable to use to perform work is known as entropy.
- Entropy Change is a phenomena that measures how disorder or randomness have changed inside a thermodynamic system.
- It has to do with how heat or enthalpy is converted during work. More unpredictability in a thermodynamic system indicates high entropy.
- Entropy is a state function, hence it is independent of the direction that the thermodynamic process takes.
- The rearranging of atoms and molecules from their initial state causes the change in entropy.
- This may result in a decrease or rise in the system's disorder or unpredictability, which will, in turn, result in a corresponding drop or increase in entropy.
To view more questions about entropy change, refer to:
brainly.com/question/4526346
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Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have
= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is
Number of moles of 
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of 
= 0.167 M
Now the ICE table :
I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,
= 
Base ionization constant, ![$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$](https://tex.z-dn.net/?f=%24K_b%20%3D%20%5Cfrac%7B%5Cleft%5BHNO_2%5Cright%5D%20%5Cleft%5BOH%5E-%20%5Cright%5D%7D%7B%5Cleft%5BNO%5E-_2%20%5Cright%5D%7D%24)
So, ![$[OH^-]=1.9054 \times 10^{-6 } \ M$](https://tex.z-dn.net/?f=%24%5BOH%5E-%5D%3D1.9054%20%5Ctimes%2010%5E%7B-6%20%7D%20%5C%20M%24)
pOH =- ![$\log[OH^-]$](https://tex.z-dn.net/?f=%24%5Clog%5BOH%5E-%5D%24)
= 
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
