No, it won't change the amount of reactants nor the products as a catalyst will only provide an alternative path where lower activation energy is needed for the process to take place.
hope this explains it
If it does, please give it a brainliest :)))
Answer: 520.04 meters.
If you convert 350 yards to meters, you get 320.04.
320.04m + 200m = 520.04m.
Take off the .04 if necessary:)
<span>The volume of water that moves past a point on a river in a given time is called the river’s DISCHARGE or FLOW RATE. The flow rate is used to measure and study bodies of water to better understand how they work. It is useful in areas such as wastewater treatment. Water flow or velocity typically increases as the the depth or hydraulic radius of the river increases.</span>
Ionic Crystal. Ionic hydrogen.postive electron transfer to atoms
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