Answer:
Dissociation Equations Worksheet
Write balanced chemical equations to represent the slight dissociation or the
complete dissociation for 1 mole of the following compounds. In the case of slight
dissociation use a double arrow and for complete dissociation use a single arrow.
Include phase notation in the equations.
1) silver chloride
2) sodium acetate
3) ammonium sulfate
4) calcium carbonate
5) potassium carbonate
6) sodium hydroxide
7) silver chlorate
8) iron(II) sulfate
9) lead(II) phosphate
10) lead(II) chromate
11) iron(III) chloride
12) calcium nitrate
13) iron(III) oxide
14) copper(II) sulfate
15) mercury(II) sulfide
16) zinc chloride
17) lead(II) acetate
18) aluminum phosphate
Solutions
1) AgCl(s) ↔ Ag+
(aq) + Cl-
(aq)
2) NaC2H3O2(s) Æ Na+
(aq) + C2H3O2
-
(aq)
3) (NH4)2SO4(s) Æ 2NH4
+
(aq) + SO4
2-(aq)
4) CaCO3(s) ↔ Ca2+(aq) + CO3
2-(aq)
5) K2CO3(s) Æ 2K+
(aq) + CO3
2-(aq)
6) NaOH(s) Æ Na+
(aq) + OH-
(aq)
7) AgClO3(s) Æ Ag+
(aq) + ClO3
-
(aq)
8) FeSO4(s) Æ Fe2+(aq) + SO4
2-(aq)
9) Pb3(PO4)2(s) ↔ 3Pb2+(aq) + 2PO4
3-(aq)
10) PbCrO4(s) ↔ Pb2+(aq) + CrO4
2-(aq)
11) FeCl3(s) Æ Fe3+(aq) + 3Cl-
(aq)
12) Ca(NO3)2(s) Æ Ca2+(aq) + 2NO3
-
(aq)
13) Fe2O3(s) ↔ 2Fe3+(aq) + 3O2-(aq)
14) CuSO4(s) Æ Cu2+(aq) + SO4
2-(aq)
15) HgS(s) ↔ Hg2+(aq) + S2-(aq)
16) ZnCl2(s) Æ Zn2+(aq) + 2Cl-
(aq)
17) Pb(C2H3O2)2(s) Æ Pb2+(aq) + C2H3O2
-
(aq)
18) AlPO4(s) ↔ Al3+(aq) + PO4
3-(aq)
Explanation:
Answer:
- <em>The mystery substance is</em> <u>C. Bromine (Br) </u>
Explanation:
<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.
<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.
Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).
The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.
The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.
Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.
You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.
Therefore, <em>the mystery substance is bromine (Br).</em>
Answer: One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.
To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl.
<h3>HOPE THIS HELPS HAVE A AWESOME DAY❤️✨</h3>
Explanation:
Answer:
C
Explanation:
A diatomic element in that list is Bromine