Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
the degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch. So I would have to go with A.
Answer: X3Y2
Explanation:
The formula is
X has a valency of 2
Y has a valency of 3
So, we interchange the valencies
Therefore, the formula is
X3Y2
The electron-group arrangement of CO₃²⁻ is trigonal planar. The molecular shape is trigonal planar, and the ideal bond angle(s) is CO₃²⁻ is 120°
<h3>What is the molecular geometry of a compound?</h3>
The position of the compound's electrons and nuclei can be seen in the molecular geometry. It demonstrates how the form of the complex is created by the interaction of electrons and nuclei.
Here, according to the VSEPR theory, the shape of the carbonate ion is trigonal planar. The carbon will be in the center.
Thus, the electron-group arrangement and the shape of the carbonate ion are trigonal planar. The bond angle will be 120°.
To learn more about molecular geometry, refer to the link:
brainly.com/question/16178099
#SPJ4
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J
(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal
(rounded to three significant figures)
95.6 cal
are needed.
