The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
given that :
molarity = 2.16 M
volume = 250 mL = 0.25 L
the molarity is given as :
molarity = number of moles / volumes in L
from this we can calculate the number of moles, we get :
number of moles of K₂Cr₂O₇ = molarity × volume
number of moles of K₂Cr₂O₇ = 2.16 × 0.25
number of moles of K₂Cr₂O₇ = 0.54 mol
Thus, The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
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7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x ]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
