Answer is: ph value of pyridine solution is 9.1.
Chemical
reaction: C₅H₅N +
H₂O → C₅H₅NH⁺ + OH⁻.<span>
c(pyridine - C</span>₅H₅N)
= 0.115M.<span>
Kb(C</span>₅H₅N)
= 1.4·10⁻⁹.
[C₅H₅NH⁺] = [OH⁻] = x; equilibrium concentration.<span>
[</span>C₅H₅N] =
0.115 M - x.
Kb = [C₅H₅NH⁺] · [OH⁻] / [C₅H₅N].
1.4·10⁻⁹ = x² / (0.115 M -x)
Solve quadratic equation: x = [OH⁻] = 0.0000127 M.<span>
pOH = -log(0.0000127 M) = 4.9</span>
<span>pH = 14 - 4.9 = 9.1.</span>
your answer will be :
B. <u>Na has a lower</u> <u>electronegativity than H</u>
because Na belongs to alkali metals which are least electronegative (most electro positive) but hydrogen is a non metal, it has higher electronegativity as compared to metals like Sodium (Na).
Answer:
58.9g of SO2 is produced
8g of oxygen remains unconsumed
Explanation:
The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:
CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)
Molar mass of CS2 = 76.139 g/mol
Molar mass of O2 = 15.99 g/mol
Molar mass of SO2 = 64.066 g/mol
Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles
Number of moles of O2 = 30g/15.999 g/mol =1.88 moles
From the chemical reaction
1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2
Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2
Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced
thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2
Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining
Or 8g of oxygen
58.9g of SO2 is produced
oxygen is the limiting
Unmmmm i’m not quite understanding brhajsnsbfbfhrndnsnsjsmsbgbtieoou that but yes
(a) chlorine gains an electron from a sodium atom