The term for a push or pull is called Force.
Answer:
IR spectroscopy can be used to identify chemical structures are present in compounds.
Explanation:
Infrared spectroscopy is a technique in organic chemistry that can be use use to identify chemical structures present in compounds because it is base on the ability of different functional groups to adsorb infrared light.
This work by shinning the infrared lights into the organic compounds to be identified, some of the frequencies of the infrared lights are adsorbed by the compounds and its identify groups of atoms and molecules in the compound.
The electronic configuration of x, y, and z.
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
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<h3>What is an electronic configuration?</h3>
Electronic configuration, also called electronic structure, is the arrangement of electrons in energy levels around an atomic nucleus.
a.
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
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b.
x= Period 1 and group18
Y=Element is in period 3 and Group 2
Z=Element is in period 3 and group 17.
c.
X, Y & Z represents Helium, magnesium, and chlorine.
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(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
pH ranges from 0 to 14, with 7 being neutral. pHs less than 7 are acidic while pHs greater than 7 are alkaline (basic).