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- Question #1565c - Socratic
The molarity of the silver nitrate solution is 0 394 mM The balanced chemical equation for this double replacement reaction looks like this AgN O3(aq) +KCl(aq) → AgCl(s) +KN O3(aq) Notice that you have a 1:1 mole ratio between silver nitrate and potassium chloride, which means that you need 1 mole of silver nitrate for every 1 mole of potassium chloride in order for the reaction to take
- Question #c79f3 - Socratic
As you know, molarity is a measure of the number of moles of solute, which in your case would be phosphoric acid, present in
- What is the mass in grams of CaCl_2 in a 3 M CaCl_2 . . . - Socratic
Molarity = mol solute liter of solution A 3 M solution of calcium chloride contains 3 moles of the solute CaCl2 in one liter of solution To convert 3 mol CaCl2 to mass in grams, multiply the given moles by the molar mass: 110 978 g mol 3mol CaCl2 × 110 978g CaCl2 1mol CaCl2 = 300 g CaCl2 (rounded to one significant figure due to 3 mol)
- If #NaOH (aq)# that was originally #0. 300 M# is left . . . - Socratic
If #"NaOH" (aq)# that was originally #"0 300 M"# is left exposed to the air and #"0 0200 M"# of #"CO"_2# dissolves in the solution, but right at that point the student catches his mistake and caps the #"NaOH"#, what is its new molarity?
- Determine the formula of A. Molarity B. Molality C. Mole . . . - Socratic
Well, "molarity" is simply the quotient "molarity"="moles of solute" "volume of solution" And "molality"="moles of solute" "kilograms of solvent" For most, dilute, AQUEOUS solutions, "molality"-="molarity" And "mole fraction" is given by the quotient chi_"the mole fraction"="Moles of component" "Total moles present in the mixture solution" And "normality" is a bit of an old-fashioned
- Site Map - Dilution Calculations Questions and Videos | Socratic
What is the molarity of a stock solution if 10 mL is diluted to 400 mL with a concentration of 0 5M?
- Question #9d6d5 - Socratic
Since molarity is defined as number of moles of solute per liter of solution, decreasing the volume of the solution while keeping the number of moles of solute constant will increase the solution's concentration That happens because you have the same number of moles of solute in a smaller volume of solution
- Question #b28e9 - Socratic
"1 4 L" The idea here is that you need to use the molarity and volume of the target solution to determine how many moles of solute it contains Since you're dealing with a dilution, you know for a fact that the number of moles of solute present in the initial solution must be equal to the number of moles of solute present in the target solution A dilution essentially decreases the
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