Answer:
Mass of NaBr produced = 23.67 g
Explanation:
Given data:
Mass of AgBr = 42.7 g
Mass of NaBr produced = ?
Solution:
Chemical equation:
2Na₂S₂O₃ + AgBr → NaBr + Na₃(Ag(S₂O₃)₂
Number of moles of AgBr:
Number of moles = mass/molar mass
Number of moles = 42.7 g/ 187.7 g/mol
Number of moles = 0.23 mol
now we will compare the moles of AgBr with NaBr.
AgBr : NaBr
1 : 1
0.23 : 0.23
Mass of NaBr:
Mass = number of moles × molar mass
Mass = 0.23 mol × 102.89 g/mol
Mass = 23.67 g
Question 4 is C. 13 I believe
you get hydrofluoric acid
Answer:
0.0025 M/min
Explanation:
The rate of a reaction can be calculated for an element, based on its stoichiometric coefficient. For a reaction:
aA + bB = cC + dD , the rate will be
r = -(1/a)x(Δ[A]/Δt) = -(1/b)x(Δ[B]/Δt) = (1/c)x(Δ[C]/Δt) = (1/d)x(Δ[D]/Δt)
Where Δ[X] is the variation of the concentration of the X compound, Δt is the time variation, and the signal of minus in the reagents compounds is because they are disappearing, so Δ[X] will be negative, and r must be positive. So, for the reaction given:
r = -(1/2)x(Δ[NO]/Δt)
r = -(1/2)x( (0.025 - 0.1)/15)
r = 0.0025 M/min