Consider four beakers. Beaker A has an aqueous solution of NaOH in which the OH - ions are represented by blue circles. Beaker B has a weak acid; HX is represented by red circles. Beaker C has a weak acid; H 2 X is represented by green circles. Beaker D has a weak acid; H 3 X is represented by yellow circles. X - ions are represented by triangles. Match the pictorial representations with the reactions given below. (a) HX ( a q ) + OH − ( a q ) → X − ( a q ) + H 2 O (b) H 2 X ( a q ) + 2OH − ( a q ) → X − ( a q ) + 2H 2 O (c) H 3 X ( a q ) + 3OH − ( a q ) → X − ( aq ) + 3H 2 O
Consider four beakers. Beaker A has an aqueous solution of NaOH in which the OH - ions are represented by blue circles. Beaker B has a weak acid; HX is represented by red circles. Beaker C has a weak acid; H 2 X is represented by green circles. Beaker D has a weak acid; H 3 X is represented by yellow circles. X - ions are represented by triangles. Match the pictorial representations with the reactions given below. (a) HX ( a q ) + OH − ( a q ) → X − ( a q ) + H 2 O (b) H 2 X ( a q ) + 2OH − ( a q ) → X − ( a q ) + 2H 2 O (c) H 3 X ( a q ) + 3OH − ( a q ) → X − ( aq ) + 3H 2 O
Solution Summary: The author explains that the correct pictorial representation needs to be matched to the acid-base reaction given below.
Consider four beakers. Beaker A has an aqueous solution of NaOH in which the OH- ions are represented by blue circles. Beaker B has a weak acid; HX is represented by red circles. Beaker C has a weak acid; H2X is represented by green circles. Beaker D has a weak acid; H3X is represented by yellow circles. X- ions are represented by triangles. Match the pictorial representations with the reactions given below.
(a)
HX
(
a
q
)
+
OH
−
(
a
q
)
→
X
−
(
a
q
)
+
H
2
O
(b)
H
2
X
(
a
q
)
+
2OH
−
(
a
q
)
→
X
−
(
a
q
)
+
2H
2
O
(c)
H
3
X
(
a
q
)
+
3OH
−
(
a
q
)
→
X
−
(
aq
)
+
3H
2
O
Use the expression below to
⚫ calculate its value and report it to the proper number of significant digits (you may need to
round your answer).
⚫ calculate the % error (or % relative error or % inherent error)
⚫ calculate the absolute error.
(20.54±0.02 × 0.254±0.003) / (3.21±0.05) =
Value:
% Error:
Absolute error: ± |
% (only 1 significant digit)
(only 1 significant digit)
In each case (more ductile, more brittle, more tough or resistant), indicate which parameter has a larger value.
parameter Elastic limit Tensile strength
more ductile
Strain at break Strength Elastic modulus
more fragile
more tough or resistant
None
Chapter 4 Solutions
Bundle: Chemistry: Principles and Reactions, 8th, Loose-Leaf + OWLv2, 1 term (6 months) Printed Access Card
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