Answer:
pH= 0.92
Explanation:
HNO3-> H^+ +NO3^-
HNO3 is a strong acid, so it fully dissociates
[HNO3] = 0.12M [H^+] = 0.12M
pH= -log[H^+]
pH=-log[.12] = 0.92
pH = 0.92
Shorter because the I-I bond in the I2 molecule is covalent whereas the bond between adjacent molecules is due to dispersion.
<h3>What is bond length?</h3>
Bond length typically falls between 0.1 and 0.2 nm. Covalent radius is the term used to describe the length of a connection formed between two comparable atoms. The order of the bond or the quantity of linked electrons between two atoms determines the bond length.
<h3>
What connection exists between the length and the strength of a bond?</h3>
The C-C bond in alkanes must be the longest since it is the weakest, and the C-C bond in alkynes must be the shortest because it appears to be the strongest, according to the data. We determined in the preceding section that the bond strength is inversely correlated to the bond length.
<h3>How can bond length be calculated in chemistry?</h3>
- The bond gets shorter as the atoms get smaller.
- The length of the bond decreases as bond multiplicity increases.
- The shorter the bond, the higher the effective nuclear charges of the bound atoms.
Learn more about bond length:
brainly.com/question/13054857
#SPJ4
Answer: The equilibrium will shift to the left.
Explanation:
If sodium acetate is added to acetic acid, the dissociation of the acetic acid is suppressed. The equilibrium position shifts to the left and the hydrogen ion concentration decreases due to decreased ionization of the acid. This common ion solution becomes less acidic than pure acetic acid.
This is so because of common ion effect. The acetate ion already present from the dissociation of acetic acid is also present in the sodium acetate. The presence of a common ion usually shifts the equilibrium position towards the left.
NaCH3CO2 (s) → Na + (aq) + CH3CO2 −(aq)
CH3CO2H(aq) ⇌ H +(aq) + CH3CO 2− (aq)
Answer:
- 122.58 kJ
Solution:
According to following equation,
<span>2 Al + 3 Cl</span>₂ <span>→ 2 AlCl</span>₃ <span> δH</span>°<span> = -1408.4 kJ
</span>
When,
54 g (2 mole) Al on reaction releases = <span>1408.4 kJ heat
So,
4.70 g of Al will release = X kJ of Heat
Solving for X,
X = (4.70 g </span>× 1408.4 kJ) ÷ 54 g
X = - 122.58 kJ