We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
Answer:
Dominant traits are always expressed when the connected allele is dominant, even if only one copy exists. Recessive traits are expressed only if both the connected alleles are recessive.
Explanation:
Freckles, cleft chin and dimples are all examples of a dominant trait.
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
∴


