Answer:
0.382g
Explanation:
Step 1: Write the reduction half-reaction
Al³⁺(aq) + 3 e⁻ ⇒ Al(s)
Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s
We will use the following relationships.
- 1 mole of electrons has a charge of 96486 C (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons pass through the cell.
- The molar mass of Al is 26.98 g/mol.
The mass of Al produced is:
P1/T1 = P2/T2
125⁰C = 398.15 k
182⁰C = 455.15 k
1.22/398.15 = p2/455.15
p2= 1.39atm
the pressure of the gas be after the temperature change is 1.39 atm
This represents a primary amine. An amine has a nitrogen group that is connected to three substituents via single bonds. The number of carbon-based substitutents determines whether it is primary, secondary, or tertiary. In this case, since 2 substitutents are just hydrogen atoms, and only one has a carbon-based skeleton, this is a primary amine.
Convert Mg to grams
1g =1000mg what about 3.91 Mg
= 3.91mg x 1g/1000mg= 3.91 x10^-3 g
moles= mass/molar mass
that is 3.91 x10^-3g /99 g/mol=3.95 x10^-5moles
concentration= moles / vol in liters
that is 3.95 x10^-5/100 x1000= 3.94 x10^-4M
equation for dissociation of CUCl= CUCl----> CU^+ +Cl^-
Ksp=(CU+)(CI-)
that is (3.95 x10^-4)(3.95 x10^-4)
Ksp= 1.56 x10^-7
