<span>Use the van't Hoff equation:
ln
(
K2
K1
)
=
Δ
HÂş
R
(
1
T1
â’
1
T2
)
ln
(
K2
7.6*10^-3
)
=
-14,200 J
8.314
(
1
298
â’
1
333
)
ln
(
K2
7.6*10^-3
)
=
â’
1708
(
0.00035
)
ln
(
K2
0.0076
)
=
â’
0.598
Apply log rule
a
=
log
b
b
a
-0.598 =
ln
(
e
â’
0.598
)
=
ln
(
1
e
0.598
)
Multiply both sides with e^0.598
K
2
e
0.598
= 0.0076
K
e
0.598
e
0.598
=
0.0076
e
0.598
K
2
=
0.0076
e
0.598
=
4.2
â‹…
10
â’
3
K2
=
4.2
â‹…
10
â’
3</span>
<h3><u>Answer;</u></h3>
<u> = 1319.45 Joules </u>
<h3><u>Explanation;</u></h3>
Kinetic energy is the energy possessed by a body in motion;
It is given by the formula; 1/2mv², where m is the mass of the body and v is the velocity.
mass = 89.5 kg and v =5.43 m/s
Therefore;
Kinetic energy = 1/2 × 89.5 × 5.43 ²
<u> = 1319.45 Joules </u>
Answer:
Value of 
is 0.090.
Explanation:
Initial molarity of 
= 
= 0.0700 M
Construct an ICE table corresponding to the combustion reaction of carbon to determine
I (M): - 0.0700 0
C (M): - -x +2x
E (M): - 0.0700-x 2x
So, ![K_{c}=\frac{[CO]^{2}}{[O_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BCO%5D%5E%7B2%7D%7D%7B%5BO_%7B2%7D%5D%7D)
, where [CO] and ![[O_{2}]](https://tex.z-dn.net/?f=%5BO_%7B2%7D%5D)
represents equilibrium concentration of CO and respectively.
Here, ![[CO]=2x=0.060](https://tex.z-dn.net/?f=%5BCO%5D%3D2x%3D0.060)
⇒x = 0.030
So, = 0.0700-x = (0.0700-0.030) = 0.040
Hence, 
Answer:
6.5 moles. 6.02 x 10. 23 atoms = 3.9 x 10. 1
