The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.
Answer:
-1273.3
Explanation:
Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.
enthalpy of formation Of CO2 = 2 X -393.5 = -787
enthalpy of formation Of C2H5OH = 2 X -277.7 = -555.4
enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + -555.4) = - 1273.3 Joules
Answer: 19.4 mL Ba(OH)2
Explanation:
H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)
At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.
0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2
Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.
0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl
HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.
0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2