Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less stable than those of the other halogens. Bonding theory, however, does allow one to propose structures for these acids and use formal charges for the evaluation of these structures. For a molecule of fluorous acid, the atoms are arranged as HOFO. (Note: In this oxyacid, the placement of fluorine is an exception to the rule of putting the more electronegative atom in a terminal position.) What is the formal charge on each of the atoms?

Use the gizmo to find the concentration of the mystery ch3cooh. use the titrant and indicator shown below perform the titration. what is the titrant volume? titrant analyte indicator titrant volume analyte concentration naoh ch3cooh phenophthalein select one: a. 20.0 ml b. 27.0 ml c. 30.0 ml d. 24.0 ml

Ineed someone to see if my answers are correct! if any are wrong let me know what the correct answers would be and how to get that answer! 1. how many moles of sodium chloride are in 28 grams od nacl? a. 265 mole naclb. 856 mole naclc. 479 mole of nacld. 1.2 mole nacl < my choice2. 734 grams of lithium sulfate (li2so4) are dissolved to make 2500 ml of solution what is rhe molaratiy? a. 2.67 mb. 4.56 mc. 3.89 m < my choiced. 1.78 m3. how many grams of cacl2 would be dissolved in 3.0 l of a 0.50 m solution of cacl2? a. 250 g cacl2 b. 166.5 g cacl2c. 113.65 g cacl2d. 98 g cacl2 < my choice4. suppose you had 58.44 g of nacl and you dissolved it in exactly 2.00 liters. the molarity if the solution would be 0.5 mtrue < my choicefalse 5. i would need 22g of naoh to make a 3.0 m solution using 250 ml of solvent.true < my choicefalse6. identify the solute: you have a .0195 m solution made from using 6.5 g of solute and 3 l of solvent. identify the solute by solving for molar weight.a. the solute is nacl because the molar weight is 58.43 g/mol < my choiceb. the solute is h2so4 because the molar weight is 98.06 g/molc. the solute is cacl2 because the molar weight is 111.11 g/mol

Ineed ! i have no clue how to do this problem.