Answers by visitors named: Fai

1ounce= 30ml 12x30ml =360ml 360ml x 9% = 32.4ml 4%X + 12Y(360ml-X)= 32.4ml 0.04X + 43.2ml + 0.12X = 32.4ml 0.076X = 10.8ml x= 142.ml 4% y = 360-142.1ml y= 218ml 12%

1.2g/ml x 1000ml x 0.37 / 36.5 = 12.16M 12.16m= 2500ml x 0.5 12.16m = 1250ml x = 1250ml/12.16 x = 103 ml This is answer.

My own calculation 1.179g/ml x 1000ml x 0.36/36.5 x = 11.62 11.62M = 4.6 L change 4600ml 11.62 M = 4600ml x 1.9 11.62M = 8740ml X= 8740ml/11.62 x= 752ml

0.1879M x 0.025L = 4.6975 mols 4.6975 x 137.9 = 647.78 changes 0.647gm I am not sure answer who tells me the right. Pls help

how to calculate PH =1.9 therefore =0.0126M I do not understand only this part about PH = 0.0126M Please tells me how to calculate the PH = 1.9 = 0.0126M Other way to calculate 0.0126Moles = PH 1.9 0.0126M x 4.85L=0.061mole 0.061mole x 36.46 = 2.22g 2.22g /0.36 = 6.16 6.16/1.179 = 5.23ml

A 0.446g sample of an unknown monoprotic acid was titrated with 0.105M KOH the molar mass is 120g/mol. What is the PKA value Who helps me to solve it for me. Please help

How to calculate PH 1.9 = 0.0126M Why? How to calculate PH 1.8 = ? How to calculate PH 20 = ? How to calculate PH 21 = ? Pls helps me

A solution of sodium cyamide nacn has a ph of 12.10. How many grams of nacn are in 425ml of solution with the same PH Who helps me to solve it for me. Step by step

I do not still understand how to calculate PH 1.9 = 0.0126M I want step by step to calculate for me. Thank you for your help.

0.79g/ml x 76.9ml = 60.7g 60.7g /46.06 = 1.3 mole 1.3/ 0.25 L = 5.2M

My calculation 0.791g/ml x 75ml /46 g/mol = 1.29M 1.29M /0.15KG x= 0.116M

Determine this volume of diluted nitric acid d = 1.11g/ml, 19% hn03. That can be prepared by diluting with water 50ml of conc hn03 d=1.42g/ml, 69.8% I do not still understand how to get 19% is .2722 , the right answer is 183.68ml