Answer:
a) pH will be 12.398
b) pH will be 4.82.
Explanation:
a) The moles of NaOH added = molarity X volume (L) = 2 X 0.01 = 0.02 moles
The total volume after addition of pure water = 0.780+0.01 = 0.79 L
The new concentration of /NaOH will be:
the [OH⁻] = 0.025
pOH = -log [OH⁻] = 1.602
pH = 14 -pOH = 12.398
b) The buffer has butanoic acid and butanoate ion.
i) Before addition of NaOH the pH will be calculated using Henderson Hassalbalch's equation:
![pH=pKa+log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
pKa=
ii) on addition of base the pH will increase.
Answer:
grams O₂ = 134 grams
Explanation:
PV = nRT => n = PV/RT
P = 8.15atm
V = 12.2 Liters
R = 0.08206L·atm/mol·K
T = 16.0°C + 273 = 289K
n = (8.15atm)(12.2L)/(0.08206L·atm/mol·K)(289K) = 4.2 moles O₂
grams O₂ = 4.2 moles O₂ x 32g/mol = 134 grams O₂
1 calorie is needed to raise 1 g of water 1 °C.
350 * 22 = 7700 calories
Answer:
If you're just looking at the Lewis Structure from the perspective of the octet rule, it does appear that the structure is correct. Dinitrogen always has a lone pair of electrons which could conceivably be used for dative bonding as you suggest. So from that perspective there appears to be nothing wrong at all - other than that it doesn't exist in nature in this way.
Explanation:
