The substance(s) or molecules at the beginning of a chemical reaction is called a chemical equation

product

reactant

Acompound is found to contain 6.1% hydrogen and 93.9% oxygen. find it’s empirical formula.

During trial 2, what allowed you to determine that aluminum was the limiting reactant? check all that apply. all of the copper dissolved. all of the aluminum dissolved. the solution turned clear. the number of grams of copper(ii) chloride used in the reaction was greater than the number of grams of aluminum. the molar ratio of copper(ii) chloride to aluminum was greater than 3: 2, the equation’s molar ratio.

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