Answer:
near an open flame
Explanation:
if your hair is not tied back near open flames in combination with most hair care products it would turn you into a living torch
A reversible reaction is a reaction that takes place in the ( formation of reactants to products and products to reactants simultaneously ). If the reaction were to ( be in dynamic equilibrium ), the rate of forward reaction would be equal to that of the reverse reaction.
hope this helps :))
Answer is: Ksp for strontium arsenate is 2.69·10⁻¹⁸.
Balanced chemical reaction (dissociation):
Sr₃(AsO₄)₂(s) → 3Sr²⁺(aq) + AsO₄³⁻(aq).
s(Sr₃(AsO₄)₂) = 0.0650 g/L.
s(Sr₃(AsO₄)₂) = 0.0650 g/L ÷ 540.7 g/mol = 1.2·10⁻⁴ mol/L.
s(Sr²⁺) = 3s(Sr₃(AsO₄)₂).
s(AsO₄³⁻) = 2s(Sr₃(AsO₄)₂).
Ksp = s(Sr²⁺)³ · s(AsO₄³⁻)².
Ksp = (3s)³ · (2s)².
Ksp = 108s⁵.
Ksp = 108 · (1.2·10⁻⁴ mol/L)⁵ = 2.69·10⁻¹⁸.
Answer is: halogens ans noble gases.
The nonmetals are divided into two categories: reactive nonmetals and noble gases.
Halogen elements are in group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). They are very reactive and easily form many compounds.
Noble gases are in group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They have very low chemical reactivity.
Answer:
0.0159m
Explanation:
9 M
Explanation:
Lead(II) chloride,
PbCl
2
, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.
Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,
K
sp
, will be established between the solid lead(II) chloride and the dissolved ions.
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
Now, the molar solubility of the compound,
s
, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.
Notice that every mole of lead(II) chloride will produce
1
mole of lead(II) cations and
2
moles of chloride anions. Use an ICE table to find the molar solubility of the solid
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
I
−
0
0
C
x
−
(+s)
(
+
2
s
)
E
x
−
s
2
s
By definition, the solubility product constant will be equal to
K
sp
=
[
Pb
2
+
]
⋅
[
Cl
−
]
2
K
sp
=
s
⋅
(
2
s
)
2
=
s
3
This means that the molar solubility of lead(II) chloride will be
4
s
3
=
1.6
⋅
10
−
5
⇒
s
= √
1.6
4
⋅
10
−
5 =
0.0159 M
