Answer:
B-2
Explanation:
In the graph you can so that two of the shapes ae fully black, that means that they are color blind, the half colored ones means they are a carrier but they aren't color blind. So two males in the offspring are color blind.
In a flame photometric analysis, salt solution is first vaporized using the heat of flame, followed by this electrons from valance shell gets excited from ground state to excited state. Followed by this de-excitation of electron bring backs electrons to ground state. This process is accompanied by emission of photon. The photon emitted is characteristic of an element, and number of photons emitted can be used for quantitative analysis.
<span>Following are the investigative question that you can answer by doing this experiment.
</span>1) What information can be obtained from the colour of flame?
2) <span>State the relationship between wavelength, frequency, and energy?
</span><span>3) Can you identify the metal present in unknown sample provided?
4) How will you identify amount of metal present in sample solution?
5) </span><span>Why do different chemicals emit light of different colour?</span><span>
</span>
(a)
pH = 4.77
; (b)
[
H
3
O
+
]
=
1.00
×
10
-4
l
mol/dm
3
; (c)
[
A
-
]
=
0.16 mol⋅dm
-3
Explanation:
(a) pH of aspirin solution
Let's write the chemical equation as
m
m
m
m
m
m
m
m
l
HA
m
+
m
H
2
O
⇌
H
3
O
+
m
+
m
l
A
-
I/mol⋅dm
-3
:
m
m
0.05
m
m
m
m
m
m
m
m
l
0
m
m
m
m
m
l
l
0
C/mol⋅dm
-3
:
m
m
l
-
x
m
m
m
m
m
m
m
m
+
x
m
l
m
m
m
l
+
x
E/mol⋅dm
-3
:
m
0.05 -
l
x
m
m
m
m
m
m
m
l
x
m
m
x
m
m
m
x
K
a
=
[
H
3
O
+
]
[
A
-
]
[
HA
]
=
x
2
0.05 -
l
x
=
3.27
×
10
-4
Check for negligibility
0.05
3.27
×
10
-4
=
153
<
400
∴
x
is not less than 5 % of the initial concentration of
[
HA
]
.
We cannot ignore it in comparison with 0.05, so we must solve a quadratic.
Then
x
2
0.05
−
x
=
3.27
×
10
-4
x
2
=
3.27
×
10
-4
(
0.05
−
x
)
=
1.635
×
10
-5
−
3.27
×
10
-4
x
x
2
+
3.27
×
10
-4
x
−
1.635
×
10
-5
=
0
x
=
1.68
×
10
-5
[
H
3
O
+
]
=
x
l
mol/L
=
1.68
×
10
-5
l
mol/L
pH
=
-log
[
H
3
O
+
]
=
-log
(
1.68
×
10
-5
)
=
4.77
(b)
[
H
3
O
+
]
at pH 4
[
H
3
O
+
]
=
10
-pH
l
mol/L
=
1.00
×
10
-4
l
mol/L
(c) Concentration of
A
-
in the buffer
We can now use the Henderson-Hasselbalch equation to calculate the
[
A
-
]
.
pH
=
p
K
a
+
log
(
[
A
-
]
[
HA
]
)
4.00
=
−
log
(
3.27
×
10
-4
)
+
log
(
[
A
-
]
0.05
)
=
3.49
+
log
(
[
A
-
]
0.05
)
log
(
[
A
-
]
0.05
)
=
4.00 - 3.49
=
0.51
[
A
-
]
0.05
=
10
0.51
=
3.24
[
A
-
]
=
0.05
×
3.24
=
0.16
The concentration of
A
-
in the buffer is 0.16 mol/L.
hope this helps :)
Answer:
V = 240.79 L
Explanation:
Given data:
Volume of butane = ?
Temperature = 293°C
Pressure = 10.934 Kpa
Mass of butane = 33.25 g
Solution:
Number of moles of butane:
Number of moles = mass/ molar mass
Number of moles = 33.25 g/ 58.12 g/mol
Number of mole s= 0.57 mol
Now we will convert the temperature and pressure units.
293 +273 = 566 K
Pressure = 10.934/101 = 0.11 atm
Volume of butane:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
V = nRT/P
V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm
V = 26.49 L/0.11
V = 240.79 L
Answer:
The mass of CH4 is 60, 29 grams.
Explanation:
We use the weight of the atoms C and H for calculate the molar mass:
Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=
Weight of CH4 =16, 042 g/mol
1molCH4-----16, 042grams
3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams
