Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of
= 207+56+192 = 455 g/mol
Formula:
Solving:
Answer:
<span>
B. 0.119 M</span>
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
Answer:
(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)
Explanation:
A buffer is a solution of a weak acid and its salt. It mitigates against changes in acidity or alkalinity of a system. A buffer maintains the pH at a constant value by switching the equilibrium concentration of the conjugate acid or conjugate base respectively.
Addition if an acid shifts the equilibrium position towards the conjugate acid side while addition of a base shifts the equilibrium position towards the conjugate base side.
The typically bald, rocky, or sandy surfaces in desert climates hold little moisture and evaporate the little rainfall they receive.
