The technique of matrix isolation involves condensing the substance to be studied with a large excess of inert gas (usually argon or nitrogen) at low temperature to form a rigid solid (the matrix). The early development of matrix isolation spectroscopy was directed primarily to the study of unstable molecules and free radicals. The ability to stabilise reactive species by trapping them in a rigid cage, thus inhibiting intermolecular interaction, is an important feature of matrix isolation. The low temperatures (typically 4-20K) also prevent the occurrence of any process with an activation energy of more than a few kJ mol-1. Apart from the stabilisation of reactive species, matrix isolation affords a number of advantages over more conventional spectroscopic techniques. The isolation of monomelic solute molecules in an inert environment reduces intermolecular interactions, resulting in a sharpening of the solute absorption compared with other condensed phases. The effect is, of course, particularly dramatic for substances that engage in hydrogen bonding. Although the technique was developed to inhibit intermolecular interactions, it has also proved of great value in studying these interactions in molecular complexes formed in matrices at higher concentrations than those required for true isolation.
3x^2 + 8x - 4 - (6x^2 - 5x + 3)....distribute thru the parenthesis
3x^2 + 8x - 4 - 6x^2 + 5x - 3 .....combine like terms
-3x^2 + 13x - 7 <==
Answer:
x=118.25
Step-by-step explanation:
3−2x=−1.5−6−226
3+−2x=−1.5+−6+−226
−2x+3=(−1.5+−6+−226)(Combine Like Terms)
−2x+3=−233.5
−2x+3=−233.5
Step 2: Subtract 3 from both sides.
−2x+3−3=−233.5−3
−2x=−236.5
Step 3: Divide both sides by -2.
−2x
−2
=
−236.5
−2
x=118.25
Answer:
1.09
Step-by-step explanation:
Check the photo
Answer:
s(5) = -12
Step-by-step explanation:
s(t) = - t³ + 7t² - 14t +8
Find the displacement during the first 5 seconds.
When t = 5
s(t) = - t³ + 7t² - 14t +8
s(5) = -5³ + 7(5)² - 14(5) + 8
= -125 + 7(25) - 70 + 8
= -125 + 175 - 70 + 8
= -12
s(5) = -12
