Answer:
To balance the reaction, we should put a 4 in front of Al
Explanation:
<u>Step 1:</u> Data given
AI+3O2->2AI2O3
<u>Step 2:</u> Balance the equation
On the right side we have 2*3 = 6 oxygen atoms
On the left side we have 3*2 = 6 oxygen atoms as well, so those are on both side balanced
On the right side we have 2*2 = 4 Al atoms
On the left side we have only 1 Al atom
We should multiply this by 4 (place a 4 in front of the Al)
4AI + 3O2 → 2AI2O3
Now al atoms are balanced on both sides.
Answer:I believe it is c.
Explanation:
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.
According to Henderson–Hasselbalch Equation,
pH = pKa + log [Lactate] / [Lactic Acid]
As,
Ka of Lactic Acid = 1.38 × 10⁻⁴
pKa = -log Ka
pKa = -log 1.38 × 10⁻⁴
pKa = 3.86
So,
pH = 3.86 + log [0.10] / [0.13]
pH = 4.74 + log 0.769
pH = 4.74 - 0.11
pH = 4.63
Answer:
That the isotope H-1 is the most abundant in nature.
Explanation:
Hello!
In this case, since the average atomic mass of an element is computed considering the mass of each isotope and the percent abundance each, for hydrogen we would set up something like this:
Moreover, since the isotope notation H-1 and H-2 means that the atomic mass of H-1 is 1 amu, that of H-2 is 2 amu and the average one is 1.0079 amu, we can infer that the most of the hydrogen in nature is H-1 as the most of it composes the average hydrogen atom.
Best regards!