The Henderson-Hasselbalch approximation is for conjugate acid-base pairs in a buffered solution. We're going to call HA a weak acid, and A- its conjugate base. The equation is as follows:
pH = pKa + log([base]/[acid]), where the brackets imply concentrations
Plugging in our symbols and the pKa value, the equation becomes:
pH = 4.874 + log([A-]/[HA])
They drill too deep and find lava
3
Explanation:
Number 1 and 2, Are good things.
if one is decreasing and the other is benefiting. Otherwise, Here as number 3.
The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
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<span>Answer: option B. 3.07 g
Explanation:
1) given reaction:
S(s) + O₂ (g) → SO(g)
2) Balanced chemical equation:
</span><span>2S(s) + O₂ (g) → 2SO(g)
3) Theoretical mole ratios:
2 mol S : 1 mol O₂ : 2 mol SO
3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa
use the ideal gas equation:
pV = nRT
with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)
=> n = pV / (RT) =
n = [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K) × 573.15 K ]
n = 0.0954 mol SO
4) proportion with the theoretical ratio S / SO
2 mol S x
-------------- = ----------------------
2 mol SO 0.0954 mol SO
=> x = 0.0954 mol S.
5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol
mass = number of moles × atomic mass
mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S
6) Therefore the answer is the option B. 3.07 g
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