If object A weighs 8 grams and has a volume of 4 mL and object B weighs 16 g and has a volume of 2 mL
B is less dense than A.
A and B have equal densities.
B is twice as dense as A.
B is four times as dense as A.

Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one species to another. oxidation states, or oxidation numbers, allow chemists to keep track of these electron transfers. in general, one element will lose electrons (oxidation), with the result that it will increase in oxidation number, and another element will gain electrons (reduction), thereby decreasing in oxidation number. the species that is oxidized is called the reducing agent or reductant. the species that is reduced is called the oxidizing agent or oxidant. to sum up: oxidation = increase in oxidation state = loss of electrons = reducing agent reduction = decrease in oxidation state = gain of electrons = oxidizing agent part a which element is oxidized in this reaction? fe2o3+3co→2fe+3co2 enter the elemental symbol. view available hint(s) is oxidized part b which element is reduced in this reaction? 2hcl+2kmno4+3h2c2o4→6co2+2mno2+2kcl +4h2o enter the elemental symbol. view available hint(s) is reduced

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

Many free radicals combine to form molecules that do not contain any unpaired electrons. the driving force for the radical–radical combination reaction is the formation of a new electron‑pair bond. consider the chemical equation. n(g)+no(g)⟶nno(g) n(g)+no(g)⟶nno(g) write lewis formulas for the reactant and product species in the chemical equation. include nonbonding electrons. n(g)n(g) select draw rings more erase select draw rings more erase select draw rings more erase n no(g)