The sun is in the middle baby :)
Answer:
C
Explanation:
The detailed solution is found in the image attached. It is necessary to note that the oxidation half equation is multiplied by three to balance electron gain and loss. This is adequately shown in the image below. Inferences are only drawn from balanced redox reaction equation hence the first step is to balance the redox reaction equation.
Answer:
The answer is "Al= 9.8% and Fe=18.0%"
Explanation:
Given:
The weight of
= 5.95g
![gFeCl_3=gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})\\\\gAlCl_3=gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})\\\\](https://tex.z-dn.net/?f=gFeCl_3%3DgFe%20%28%5Cfrac%7BMw%5C%20FeCL_3%7D%7B%5Ctext%7Batomic%20weight%20of%20Fe%7D%7D%29%5C%5C%5C%5CgAlCl_3%3DgAl%20%28%5Cfrac%7BMw%5C%20AlCL_3%7D%7B%5Ctext%7Batomic%20weight%20of%20Al%7D%7D%29%5C%5C%5C%5C)
![\to a = FeCl_3+AlCl_3\\\\\to a=x+y \\\\ \to a= 5.95](https://tex.z-dn.net/?f=%5Cto%20a%20%3D%20FeCl_3%2BAlCl_3%5C%5C%5C%5C%5Cto%20%20a%3Dx%2By%20%5C%5C%5C%5C%20%20%5Cto%20a%3D%205.95)
![\to a =x \ gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})+ y \ gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})\\\\](https://tex.z-dn.net/?f=%5Cto%20a%20%3Dx%20%5C%20gFe%20%28%5Cfrac%7BMw%5C%20FeCL_3%7D%7B%5Ctext%7Batomic%20weight%20of%20Fe%7D%7D%29%2B%20y%20%5C%20gAl%20%28%5Cfrac%7BMw%5C%20AlCL_3%7D%7B%5Ctext%7Batomic%20weight%20of%20Al%7D%7D%29%5C%5C%5C%5C)
![\to x (\frac{162.2}{55.85})+ y (\frac{133.34}{26.98})= 5.95\\\\\to 2.90x+4.94y=5.95\\\\\ similarly \ for \ oxidies:\\\\\to 143x+1.89y=2.62\\\\\to x= 1.07 \ \ \ and \ \ y= 0.58\\\\\to \ Al \% = \frac{0.58}{5.95} \times 100= \bold{9.8} \%\\\\\to \ Fe \% = \frac{1.07}{5.95} \times 100= \bold{18.0} \%](https://tex.z-dn.net/?f=%5Cto%20x%20%28%5Cfrac%7B162.2%7D%7B55.85%7D%29%2B%20y%20%28%5Cfrac%7B133.34%7D%7B26.98%7D%29%3D%205.95%5C%5C%5C%5C%5Cto%202.90x%2B4.94y%3D5.95%5C%5C%5C%5C%5C%20similarly%20%5C%20for%20%5C%20oxidies%3A%5C%5C%5C%5C%5Cto%20143x%2B1.89y%3D2.62%5C%5C%5C%5C%5Cto%20x%3D%201.07%20%5C%20%5C%20%5C%20and%20%5C%20%5C%20y%3D%200.58%5C%5C%5C%5C%5Cto%20%5C%20Al%20%5C%25%20%3D%20%5Cfrac%7B0.58%7D%7B5.95%7D%20%5Ctimes%20100%3D%20%5Cbold%7B9.8%7D%20%5C%25%5C%5C%5C%5C%5Cto%20%5C%20Fe%20%5C%25%20%3D%20%5Cfrac%7B1.07%7D%7B5.95%7D%20%5Ctimes%20100%3D%20%5Cbold%7B18.0%7D%20%5C%25)
Answer:
In water PH+Poh=14. Ph=14-pOH=14-5=9
Explanation:sorry its so long hope this helps
O
H
=
−
log
10
[
H
O
−
]
by definition.
Thus
p
O
H
=
−
log
10
(
1
×
10
−
5
)
=
−
(
−
5
)
=
5
p
H
+
p
O
H
=
14
p
H
=
14
−
p
O
H
=
14
−
5
=
9
When we write
log
a
b
=
c
, we ask to what power we raise the base,
a
, to get
b
; here
a
c
=
b
. The normal bases are
10
,
(common logarithms)
, and
e
,
(natural logarithms)
.
Thus when we write
log
10
(
10
−
5
)
, we are asking to what power we raise
10
to get
10
−
5
. Now clearly the answer is
−
5
, i.e.
log
10
(
10
−
5
)
=
−
5
, alternatively
log
10
(
10
5
)
=
+
5
.
What are
log
10
(
100
)
,
log
10
(
1000
)
,
log
10
(
1000000
)
?
?
You shouldn't need a calculator, but use one if you don't see it straight off.
A is correct vegetables cannot be like acids or alkaline because they foods