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
I do not know i am doing this for the points
b its plasma membrane probably
Formula is (CH3)2CHCH2C(CH3)3 and the name is 2,2,4-trimethylpentane
Ksp for FeCO₃ is 3.07 x 10⁻¹¹
Ksp for MgCO₃ is 6.82 x 10⁻⁶
So Fe²⁺ will precipitated first as solubility product of FeCO₃ is lower than solubility product of MgCO₃
Ksp FeCO₃ = [Fe²⁺][CO₃²⁻]
3.07 x 10⁻¹¹ = (2.3 x 10⁻²) [CO₃²⁻]
[CO₃²⁻] = 1.33 x 10⁻⁹ M to precipitate Fe²⁺ ions
Ksp MgCO₃ = [Mg²⁺][CO₃²⁻]
6.82 x 10⁻⁶ = (2.0 x 10⁻²) [CO₃²⁻]
[CO₃²⁻] = 3.4 x 10⁻⁴ M to precipitate the Mg²⁺ ions
Ksp FeCO₃ = [Fe²⁺][CO₃²⁻]
3.07 x 10⁻¹¹ = [Fe²⁺] (3.4 x 10⁻⁴)
[Fe²⁺] = 9 x 10⁻⁸ M when Mg²⁺ ions precipitate
