Answer:
A student titrates 20.0mL of 1.0M NaOH with 2.0M formic acid, HCO2H (Ka=1.8x10-4). Formic ... How much formic acid is necessary to reach the equivalence point? a. ... At the equivalence point, is the solution acidic, basic or neutral? ... By adding a small amount of water to the beaker but not dissolving all of the solid. 11.
Explanation:
Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
Answer:
0.0011 mol/L.s
Explanation:
The average rate of disappearing of the reagent is the variation of the concentration of it divided by the time that this variation is being measured. The reaction rate, is proportional to the coefficient of the substance, so, for a generic reaction:
aA + bB --> cC + dD
rate = -(1/a)Δ[A]/Δt = -(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δdt = (1/d)Δ[D]/dt
The minus sign is because of the reagent is desapering, so:
rate = -(1/2)*(0.0209 - 0.0300)/(10 - 6)
rate = 0.0011 mol/L.s