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.
An atom is the smallest particle of an element that can take part in a chemical reaction.
An atom is made up of energy levels that contain electrons which are negatively charged and the nucleus which contains neutrons and protons that are negatively charge .
Due the positive charge of the nucleus of an atom, an atom always want to attract its electrons and keep them near it however it weakly attracts the other electrons of a nearby atom.
Answer:
5.20 grams of Br₂
Explanation:
From our previous knowledge;
We understand that:
The number of moles of a given element = mass of the element divided by its molar mass.
Mathematically:
![\mathbf{no \ of \ moles =\dfrac{ mass}{ molar \ mass}}](https://tex.z-dn.net/?f=%5Cmathbf%7Bno%20%5C%20%20of%20%5C%20%20moles%20%3D%5Cdfrac%7B%20mass%7D%7B%20%20molar%20%20%5C%20mass%7D%7D)
From the given information, let's assume that the 0.065 moles of liquid -bromine partake in the reaction.
From the periodic table, the molar mass of Bromine is = 79.9 g/mol
As such, the mass of liquid that partakes is calculated as:
0.065 mol = mass/ 79.9 g/mol
mass = 0.065 mol × 79.9 g/mol
mass of liquid that partakes in the reaction = 5.20 grams of Br₂
Answer:
4.1 moles
Explanation:
Applying
PV = nRT................ equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT.............. Equation 2
From the question,
Given: V = 35 L , P = 2.8 atm, T = 15 °C = (15+273) = 288 K, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (35×2.8)/(0.083×288)
n = 4.1 moles
The reaction involved in present case is:
Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)
In terms of oxidation and reduction, the reaction can be shown at
Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)
From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.
From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.
Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.