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8) ΔTb = i*Kb*m
m is molality
Since same number of mol is added to same amount of water in both cases
m will be same for both
is 1 for glucose since it is covalent compound
is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻
So, ΔTb will be 4 times in aluminum nitrate case
So, boiling point will change by 4ºC
9) use Q = m* L
L = heat of vaporization so:
T1=T2=100ºC
5.40 * 1000 => 5400 cal/g
Q = 5400 / 540
Q = 10 grams
Hope that thlps!
Onization energy is the energy required to lose an electron and form an ion. The stronger is the attraction of the atom and the electron the higher the ionization energy, and the weaker is the attraction of the atom and the electron the higher the ionization energy. This leads to a clear trend in the periodic table. Given that the larger the atom the weaker the attraction of the atom to the valence electrons, the easier they will be released, and the lower the ionization energy. This is, as you go downward in a group, the ionization energy decreases. So, the element at the top of the group will exhibit the largest ionization energy. <span>Therefore, the answer is that of the four elements of group 7A, fluorine will have the largest first ionization energy.</span>
The chemical equation would be:
2NO(g) + O2(g) --> 2NO2 (g)
<span>At equilibrium state, the partial pressure of the gases would be as follows : </span>
<span>NO = 522 - 2x </span>
<span>O2 = 421 - x </span>
<span>NO2 = 2x </span>
<span>- - - - - - - - - - - - -</span>
<span>943 - x = 748 </span>
<span>x = 195</span>
Calculating for Kp,
<span>Kp = (NO2)^2/ ((NO)^2 * (O2)) </span>
<span>Kp = (2 * 195)^2/ ((522 - 2 * 195)^2 * (421 - 195)) </span>
<span>Kp = 0.0386 </span>
