6.3072 g >>molarity = moles of solute/volume of solution (in litres) 0.45 m = n/0.4 l n = 0.45 m × 0.4 l = 0.18 mol you need 0.18 mol of nh_4oh molar mass of nh_4oh is 35.04 g/mol mass of solute = 0.18 cancelmol × 35.04 g/cancelmol = 6.3072 g We want the standard enthalpy of formation for ca (oh)_2 If naoh is added, then we have increased the concentration of hydroxide ions
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So this is a propanol derivative
Both names seem to be unambiguous.
Each water molecule contains two hydrogens, so we multiply water by two to get The acid in excess is then titrated with n aoh (aq) of known concentration.we can thus get back to the concentration or molar quantity of m (oh)2.as it stands the question (and answer) are hypothetical. If 575 grams of frozen ethonal are warmed to the freezing point of ethanol, how much heat is required to melt all of this frozen ethanol, ch3ch2oh, at the freezing point The molar heat of vaporization of ethanol= 4.95 kj/mol
