Answer:
Using the molarity of the solution
Explanation:
The concentration of the two solutions can be compared by the use of the number of moles of solute present in each of the solutions. The solution with a higher molarity will be concentrated while the solution with a lower molarity will be dilute.
Answer:
B) Eating a antacid to calm your upset stomach.
Explanation:
Answer:
19.9 mol
Explanation:
Use <em>Avogadro’s number</em> to convert formula units of CaI₂ to moles of CaI₂.
1 mol CaI₂ ≡ 6.022 × 10²³ formula units CaI₂
Moles of CaI₂ = 1.20 × 10²⁵ × (1 /6.022 × 10²³)
Moles of CaI₂ = 19.9 mol
Reaction is NH4OH <-> NH4+ OH- (note this is reversible)
Draw up an ICE table
Let x be equilibrium conc of OH- assume init conc of OH is 0M and init conc of NH4+ is 0M also. Init conc of NH4OH is 0.1M so equilibrium conc will be 0.1-x.
%dissociation = x/0.1-x * 100%
1 = 100x/0.1-x
0.1-x = 100x
101x = 0.1
x = 0.0009901
pOH = -log(0.0009901) = 3.00
Answer:
160.3g
Explanation:
We know the equation:
No of moles = mass ÷ Mass of element
We need to find the mass, so make mass the subject of the formula.
Mass = No. of moles × mass of element
Mass = 5 × 32.06
Mass = 160.3g
