Answer:
The concentration of acetic acid is 8.36 M
Explanation:
Step 1: Data given
Volume of acetic acid = 1.00 mL = 0.001 L
Volume of NaOH = 32.40 mL = 0.03240 L
Molarity of NaOH = 0.258 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate the concentration of the acetic acid
b*Ca*Va = a*Cb*Vb
⇒with b = the coefficient of NaOH = 1
⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED
⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L
⇒with a = the coefficient of CH3COOH = 1
⇒with Cb = the concentration of NaOH = 0.258 M
⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L
Ca * 0.001 L = 0.258 * 0.03240
Ca = 8.36 M
The concentration of acetic acid is 8.36 M
Answer:
437 g
Explanation:
as mass = density × volume
Answer:
1.8g
Explanation:
Initial volume = 43.5ml
Final volume = 49.4ml
Mass = 10.88g
Density = ?
Volume = Final volume - initial volume
= 49.4 - 43.5
= 5.9ml
Density = Mass/volume
Density = 10.88/5.9
= 1.8g/ml
Answer:
Choice A: approximately
.
Explanation:
Note that the unit of concentration,
, typically refers to moles per liter (that is:
.)
On the other hand, the volume of the two solutions in this question are apparently given in
, which is the same as
(that is:
.) Convert the unit of volume to liters:
.
.
Calculate the number of moles of
formula units in that
of the
solution:
.
Note that
(sulfuric acid) is a diprotic acid. When one mole of
completely dissolves in water, two moles of
ions will be released.
On the other hand,
(sodium hydroxide) is a monoprotic base. When one mole of
formula units completely dissolve in water, only one mole of
ions will be released.
ions and
ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid
dissolves in water completely, it will take two moles of
to neutralize that two moles of
produced. On the other hand, two moles formula units of the monoprotic base
will be required to produce that two moles of
. Therefore,
and
formula units would neutralize each other at a two-to-one ratio.
.
.
Previous calculations show that
of
was produced. Calculate the number of moles of
formula units required to neutralize that
.
Calculate the concentration of a
solution that contains exactly
of
formula units:
.
The micromoles of mercury(II) iodide : 0.013 μ moles
<h3>Further explanation</h3>
Given
215.0mL of a 6.0x10⁻⁵mmol/L HgI₂
Required
micromoles of HgI₂
Solution
Molarity(M) = moles of solute per liters of solution
Can be formulated :
M = n : V
n = moles
V = volume of solution
V = 215 mL = 0.215 L
so moles of solution :
n = M x V
n = 6.10 mmol/L x 0.215 L
n = 1.312 . 10⁻⁵ mmol
mmol = 10³ micromol
so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles