Answer : The given compound belongs to ether and alcohol.
Explanation :
The chemical formula of the given compound is, 
First we have to calculate the degree of unsaturation.
Formula used:
Degree of unsaturation = 
where,
C = number of carbon
H = number of hydrogen
N = number of nitrogen
X = number of halogen
Degree of unsaturation = 
The degree of unsaturation is, 0 that means there is no double or triple bond in the compound only single bond is present between the atoms.
Thus, the given compound belongs to ether and alcohol.
Answer:
2.28 × 10^-3 mol/L
Explanation:
The equation for the equilibrium is
CN^- + H2O ⇌ HCN + OH^-
Ka = 4.9 × 10^-10
KaKb = Kw
4.9 × 10^-10 Kb = 1.00 × 10^-14
Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5
Now, we can set up an ICE table
CN^- + H2O ⇌ HCN + OH^-
I/(mol/L) 0.255 0 0
C/(mol/L) -x +x +x
E/(mol/L) 0.255 - x x x
Ka = x^2/(0.255 - x) = 2.05 × 10^-5
Check for negligibility
0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255
x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6
x = sqrt(5.20 × 10^-6) = 2.28 × 10^-3
[OH^-] = x mol/L = 2.28 × 10^-3 mol/L
Answer:
0.184 atm
Explanation:
The ideal gas equation is:
PV = nRT
Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.
So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.
The decomposition occurs:
N₂O₃(g) ⇄ NO₂(g) + NO(g)
So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.
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