So, you're given 75.0 grams of C2H2Cl4 and you want to find how many L of Cl2.
Reaction: 2Cl2 + C2H2 --> C2H2Cl4
Understand the information you need to fill your conversion factor.
You'll want to know:
how many grams are in 1 mol C2H2Cl4 (167.84 g)[added atomic weight from Periodic Table]
mole to mole ratio of Cl2 and C2H2Cl4 (2:1)[given in reaction]
how many liters are in a mole (22.4 L/mol)[universal]
Then, you want to set up your conversion factor.
75.0 g C2H2Cl4 x ( 1 mol C2H2Cl4/ 167.84 g C2H2Cl4 ) x (2 mol Cl2/1 mol C2H2Cl4) x (22.4 L/ 1 mol) = ?
124,9 g grams of ammonium carbonate are needed.
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
Molarity = moles of solute/Liters of solution 0.320 M CaCl2 = moles CaCl2/4.5 Liters = 1.44 moles of CaCl2 1.44 moles CaCl2 (110.978 grams/ 1 mole CaCl2) = 159.81 grams needed so, considering the sigi figis, 160 grams needed.
The question was asked in a roundabout way, but I think you mean that if a saturated solution of sodium chloride (salt) and water contained 2 grams of sodium chloride and 5 liters of water, than how many grams of salt would be needed to make a saturated solution with 28 liters of water. First off, the one unit you gave me was mass, and the other was volume, so technically I couldn't discover how many grams of salt would be needed. However, if I can safely assume that the water in the first and second example were under the same conditions, than the answer would be 11.2 grams. Next time you submit a question be sure you are clear about what you're asking, otherwise you may receive an answer that doesn't adequately address the question you meant to ask.
C2H4O2 Find moles first. Molarity = moles of solute/Liters of solution moles C2H4O2 = molarity/Liters solution moles C2H4O2 = 0.1 M/10 L = 0.01 moles C2H4O2 -----------------------------------now, molar mass time moles 0.01 moles C2H4O2 (60.052 grams/1 mole C2H4O2) = 0.60 grams acetic acid needed --------------------------------------------
newtest3
124,9 g grams of ammonium carbonate are needed.
I suppose that this solution doesn't exist.
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
Molarity = moles of solute/Liters of solution 0.320 M CaCl2 = moles CaCl2/4.5 Liters = 1.44 moles of CaCl2 1.44 moles CaCl2 (110.978 grams/ 1 mole CaCl2) = 159.81 grams needed so, considering the sigi figis, 160 grams needed.
Molarity = moles of solute/liters of solution 1/10^pH = Molarity ( or 10^ - pH ) = 1 X 10^-4 M ---------------------- find moles CH3COOH moles of solute = 1 X 10^-4 M/3.4 liters = 2.94 X 10^-5 moles CH3COOH (60.052 grams/1 mole CH3COOH) = 0.0018 grams CH3COOH needed ------------------------------------------------- ( 1.8 X 10^- 3 grams )
You cannot convert grams to liters. 500g = 17.636981 ounces.
The question was asked in a roundabout way, but I think you mean that if a saturated solution of sodium chloride (salt) and water contained 2 grams of sodium chloride and 5 liters of water, than how many grams of salt would be needed to make a saturated solution with 28 liters of water. First off, the one unit you gave me was mass, and the other was volume, so technically I couldn't discover how many grams of salt would be needed. However, if I can safely assume that the water in the first and second example were under the same conditions, than the answer would be 11.2 grams. Next time you submit a question be sure you are clear about what you're asking, otherwise you may receive an answer that doesn't adequately address the question you meant to ask.
Molarity = moles of solute/Liters of solution ( 918 ml = 0.918 liters )rearranged algebraically,moles of solute = Liters of solution * Molaritymoles of NaOH = (0.918 l)(0.4922 M)= 0.45184 moles NaOH=======================so,0.45184 moles NaOH (39.998 grams/1 mole NaOH)= 18.1 grams sodium hydroxide needed============================
A US pint is 473 grams. - An Imperial pint is 568 grams
C2H4O2 Find moles first. Molarity = moles of solute/Liters of solution moles C2H4O2 = molarity/Liters solution moles C2H4O2 = 0.1 M/10 L = 0.01 moles C2H4O2 -----------------------------------now, molar mass time moles 0.01 moles C2H4O2 (60.052 grams/1 mole C2H4O2) = 0.60 grams acetic acid needed --------------------------------------------
6.5 - 7 bottles