There are approximately 4 resonance structures that can be drawn for N205 (no N-N bond) (minimal formal charge).
Answer:
Explanation:
Nickel is the only metal listed. Metal elements have metallic bonds between the metal atoms. Positive metal ions in a sea of electrons.
If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.
<h3>What is Charles's law?</h3>
Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.
It is expressed as;
V₁/T₁ = V₂/T₂
Given the data in the question;
- Initial temperature of gas T₁ = 100K
- Initial volume of gas V₁ = 300mL
- Final temperature T₂ = 200K
V₁/T₁ = V₂/T₂
V₂ = V₁T₂ / T₁
V₂ = ( 300mL × 200K ) / 100K
V₂ = 60000mLK / 100K
V₂ = 600mL
Therefore, if the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.
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Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.