-so, now I find the #moles using Br's mm (79.904 g/mol), and get 1.95 mol So there is my answer, 1.95 mol/L is the molarity in the 5% solution. Does this seem correct? It really comes down the whole 5% Bromine by volume assumption that I made...I think. Any help appreciated...
Moles and Molarity For acid-base chemistry purposes, it is more appropriate to measure solute concentration in moles, or individual particles (e.g., atoms, molecules), per unit volume rather than mass per unit volume. This is because atoms react with each other in known proportions in a way...
How do you solve for mole fraction, density and %w/w, when you are given Cr2(SO4)3, with molarity of 1.26 M and molality of 1.37 m? Concentration Units: Both concentration units of molarity and molality are expressed in mole...
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Complete the following table: | |Mass|Mol|Volume| |Solute|Solute|Solute|Solution|Molarity |KNO3|23.5 g|___|135.0 mL|___ |NaHCO3|___|___|200.0 mL|0.140 M |C12H22O11|56.38 g|___|___|_0.1 Complete the following table: C a t i o n A n i o n N a m e F o r m u l ...
Complete the following table: | |Mass|Mol|Volume| |Solute|Solute|Solute|Solution|Molarity |KNO3|23.5 g|___|135.0 mL|___ |NaHCO3|___|___|200.0 mL|0.140 M |C12H22O11|56.38 g|___|___|_0.1 Complete the table below. F o r m u l a C o m p o u n d n a m e I o n ...