Δu of , in kj/kg, as it isto k, (a)as a of , (b) at , (c) at .

U=3/2 RT

T=700-500=200K
R(is a constant)=8.314J/(mol*K)
U=3/2 *8.314*200=2494.2 J/mol
We need to convert mol to kg

n=m/M for hydrogen M=2
U=2494.2*2=4988.4 J/g=4988.4 kj/kg

Answer two atoms of oxygen;

Following are the  advantages of the parametric equations: 1) they can be used to graph curves that are not functions, like the unit circle.2) the parameter can be used to represent the quantity that is  useful to  provide us with better information about the graph.3)  it is useful to convert between rectangular equations and parametric equations.with specific reference to chemistry,  parametric equations are employ for solving hydrogen and hydrogen-like ions, quantum mechanically. herein, cartesian coordinates are converted in spherical polar coordinates. the solution obtained, when plotted on spherical polar graph paper, provides visualization of atomic orbital such as s,p,d and f.