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.
Explanation:
Chromosomal DNA is packaged inside microscopic nuclei with the help of histones. These are positively-charged proteins that strongly adhere to negatively-charged DNA and form complexes called nucleosomes. Each nuclesome is composed of DNA wound 1.65 times around eight histone proteins.
Th first one is Iron and the second one is Xenon
Answer:
Dispersion forces
Dipole-Dipole interaction
Explanation:
The London dispersion force refers to the temporary attractive force that acts between the electrons in two adjacent atoms when the atoms develop temporary dipoles. Dispersion forces act between any two molecules even when other intermolecular forces are in operation as long as the molecules are in close proximity to each other.
Now, CO is polar and the HCN is also polar molecule. Hence, dipole - dipole interaction forces are also in operation and acts between the two molecules in close proximity to each other.
pH = 2.1
Let 
resembles the acid; as a weak acid (a small value of 
) 
would partially dissociate to produce protons 
and 
, its conjugate base. Let the final proton concentration (i.e., ![[H^{+}]](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%20)
) be 
. (Apparently 
) Construct the following RICE table:
By definition, (all concentrations are under equilibrium condition)
![\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right](https://tex.z-dn.net/?f=%20%20%5Cleft%5Cbegin%7Barray%7D%7Bccc%7DK_%7Ba%7D%26%3D%26%5BH%5E%7B%2B%7D%5D%20%5Ccdot%20%5BA%5E%7B-%7D%5D%20%2F%20%5BHA%5D%5C%5C%26%3D%26x%5E%7B2%7D%20%2F%280.14%20-%20x%29%5Cend%7Barray%7D%5Cright%20)
It is given that
Equating and simplifying the two expressions gives a quadratic equation; solve the equation for gives:
The pH of a solutions equals the opposite of the logarithm of its proton concentration to base 10; thus for this particular solution
![\text{pH} = -\text{ln(}[H^{+}]\text{)} / \text{ln(}10\text{)} = 2.1](https://tex.z-dn.net/?f=%20%5Ctext%7BpH%7D%20%3D%20-%5Ctext%7Bln%28%7D%5BH%5E%7B%2B%7D%5D%5Ctext%7B%29%7D%20%2F%20%5Ctext%7Bln%28%7D10%5Ctext%7B%29%7D%20%3D%202.1%20)
