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OpenStudy (anonymous):
Indicate the concentration of each ion present in the solution formed by mixing 2.38 g of NaCl in 50.0 mL of 0.400 M CaCl2 solution.
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OpenStudy (anonymous):
well first here is would covert 2.38g sodium chloride into moles....can you do this?
OpenStudy (anonymous):
yes. 2.38g/58.44g=4.07*10^-2 moles
OpenStudy (anonymous):
nice....now since we have .05L of .4M CaCl2 then how many moles of CaCl2 do we have?
OpenStudy (anonymous):
.02 moles?
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