Answer:
A) 1059 J/mol
B) 17,920 J/mol
Explanation:
Given that:
Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4
R (constant) = 8.314
We know that:
We can determine 
from above if we make the subject of the formula as:
A).
The formula for calculating change in internal energy is given as:
If we integrate above data into the equation; it implies that:
Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.
B).
If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.
then T = 273 K & T2 = 1073 K
∴
A) Temperature does not affect reaction rate
Answer:
kb = 2,0x10⁻⁵
Explanation:
The ka for HCN is:
HCN ⇄ H⁺ + CN⁻; ka = 4,9x10⁻¹⁰ <em>(1)</em>
The inverse reaction has an equilibrium constant of:
H⁺ + CN⁻ ⇄ HCN k = 1/4,9x10⁻¹⁰ = 2,0x10⁹ <em>(2)</em>
As the equilibrium of the water is:
H₂O ⇄ H⁺ + OH⁻; kw = 1x10⁻¹⁴ <em>(3)</em>
The sum of (2) + (3) gives:
H₂O + CN⁻ ⇄ HCN + OH⁻; kb = kw×k = 1x10⁻¹⁴×2,0x10⁹ =
2,0x10⁻⁶; <em>kb = 2,0x10⁻⁵</em>
<em />
<em>-In fact, the general formula to convert from ka to kb is:</em>
<em>kb = kw / ka-</em>
<em />
I hope it helps!
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
Answer:
The density of Lithium β is 0.5798 g/cm³
Explanation:
For a face centered cubic (FCC) structure, there are total number of 4 atoms in the unit cell.
we need to calculate the mass of these atoms because density is mass per unit volume.
Atomic mass of Lithium is 6.94 g/mol
Then we calculate the mass of four atoms;
⇒next, we estimate the volume of the unit cell in cubic centimeter
given the edge length or lattice constant a = 0.43nm
a = 0.43nm = 0.43 X 10⁻⁹ m = 0.43 X 10⁻⁹ X 10² cm = 4.3 X 10⁻⁸cm
Volume of the unit cell = a³ = (4.3 X 10⁻⁸cm)³ = 7.9507 X 10⁻²³ cm³
⇒Finally, we calculate the density of Lithium β
Density = mass/volume
Density = (4.6097 X 10⁻²³ g)/(7.9507 X 10⁻²³ cm³)
Density = 0.5798 g/cm³
