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Author: Topic: 0.1 M HCL solution (Read 15905 times) 0 Members and 1 Guest are viewing this topic.
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Best Answer: Given data H0 = 75.0 kJ Pressure, P = 43.0 atm = 43 atm * 101325 Pa/1 atm = 4356975 Pa = 4356975 N / m2 Volume change, V = 2.0 L  5.0 L = 3.0 L ...
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A 20.00 mL sample of 0.009873 mol/L oxalic acid was titrated with potassium permanganate. If 32.33 mL of a KMnO4 solution was required to reach the pink end point ...
Thanks for your answer Dr Buzz.. I calculated that at 1st but I need the HCl amount in mL. So i suspect im meant to use the V1 x M1 = V2 x M2 equation..
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Best Answer: a. 7.5 mL b/c .10 M means .10 mol/1L which in proportion is 0.10mol/1000mL=xmol/150mL which then means there are .015 moles of NaCl Then we set up a new ...
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Best Answer: M1V1=M2V2 14.8 * V1 = 0.25*100 V1 = 0.25*100/14.8 V1 = 1.689ml M1V1= M2V2 14.8*10 = M2*250 M2 = 14.8*10/250 M2 = 0.592M
Best Answer: molarity= number of moles/ volume (in L) 3.0 ml of the 0.25% solution contain 3 * 0.25 / 100 / 180 = 4.17 e5 mol divide this by 0.1 L to get 4.17 * 10 ...
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Best Answer: M1V1=M2V2 14.8 * V1 = 0.25*100 V1 = 0.25*100/14.8 V1 = 1.689ml M1V1= M2V2 14.8*10 = M2*250 M2 = 14.8*10/250 M2 = 0.592M