How many moles are found in 5.00 g of NaCl (58.44 g/mol)?
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How many grams of NaCl (1 molar mass = 58.44 g/mol) are required to make 350 mL of a 1.25 M NaCl solution
Suppose 5.00 g of NaCl (FW = 58.44) and 5.00 g of Mg(NO3)2 (FW= 148.3) are added to 500.0 g of H2O at 25 °C. Find the mean activity coefficient of the Mg2+ and NO3 ions. loo X -A 1 .2-1 I'? 5g /58.44g/mol 0.085 SGnol
How much NaCl (58.44 g/mol) would be needed to make 500 ml of a 2mM NaCl solution?
Given 15.0 g of NaCl (MM of NaCl = 58.44 g/mol) is dissolved in 100.0 g of water: a. Determine the mass percent of this solution. b. Assume the density of the solution is 1.00 g/ml, calculate the molarity of the solution.
What is molarity of a solution prepared by dissolving 4.65 g NaCl (FM = 58.44 g/mol) in enough water to make 275 mL of NaCl solution?
Calculate the Kf of a solvent when 50.0 g of NaCl (58.44 g/mol) is dissolved in 250.g of solvent and lowers the freezing point by 6.36°C. (Assume that the solute completely dissociates.) O 0.68 °C/m O 1.36 °C/m O 1.86 °C/m O 0.93 °C/m
A 1.000 L solution contains 0.5844 g NaCl (molar mass 58.44 g/mol) and 7.455 g KCl (molar mass 74.55 g/mol), both strong electrolytes. The concentration of Cl− ions in this solution is (A) 0.1000 mol/L (B) 0.1100 mol/L (C) 0.2000 mol/L (D) 1.000 mol/L (E) 1.100 mol/L
When 1.00g of NaCl (MM 58.44 g/mol) is dissolved in 50.00g of water, the temperature drops by 0.319degree C. This temperature change is only detectable by the most sensitive thermometers. If this process takes place in a very well insulated calorimeter, no calorimeter constant is needed. a. Calculate the enthalpy change per gram of NaCl. b. Calculate the enthalpy change per mole of NaCl.
1. How many moles of NaCl (s) can be formed from 32 moles of Cl2 (g) reacting with an excess of Na (s)? Na (s) + Cl2 (g) -> NaCl (s) 2. How many moles of H2CO, can be formed from 2 moles of HCl reacting with an excess of Na2CO3? HCI + Na2CO3 -> NaCl + H2CO3
lestion 22 of 23 > How many moles of NaCl, if mixed with excess Pb2+ ions in solution, would be needed to form 13.1 g of PbCI,? moles Naci: mol