Answer:
c. 9.94 g
Explanation:
From the question,
Using
mt = m₀e⁻kt.................... Equation 1
Where mt = mass of the leaf remaining in the bag, m₀ = original mass of leave that was placed in the bag, k = decay constant, t = time.
Given: m₀ = 33 g, k = 0.04, t = 30 days.
Substitute into equation 1
mt = 33(e⁻(0.04ˣ30))
mt = 33e⁻¹²/¹⁰
mt = 33/e¹²/¹⁰
mt = 33/3.320
mt = 9.94 g.
Hence the right answer is c. 9.94 g
Answer:
2KI → 2K + I₂
Explanation:
(not my definition, but form a trustworthy source)
The definition of a decomposition reaction is:
a chemical reaction in which one reactant breaks down into two or more products.
In this case, potassium iodide will decompose into potassium and iodine.
2KI → 2K + 2I
2 Potassium and 2 Iodine on both sides, so equation is balanced.
Or, it can be written as 2KI → 2K + I₂
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
Answer:
d
Explanation:
nCl2 = 0,1 -> nRCI = 0,2
M muéi = R + 35,5 = 14,9/0,2
-> R = 39: R la K
