HELP!!!!How would you prepare the following aqueous solutions?
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Answer:
394 g of 19.0 mass %HCl Weight of HCl = 19*394/100=74.86 g of HCl Weight of water= 394-74.86= 319.14 g of water To able to have 74.86 g of HCl, the mass of solution 30.8 mass %HCl is = 74.86/30.8 * 100=243.05 g which contain : 243.05-74.86= 168.19 g of water We needed 319.14 g of water to produce desire product, so the water need to add in is 319.14-168.19= 150.95 g Final Answer: To make 394g of 19.0 mass % HCl, we add 150.95 g of water in 243.05 g of 30.8 mass %HCl
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394 g of 19.0 mass %HCl Weight of HCl = 19*394/100=74.86 g of HCl Weight of water= 394-74.86= 319.14 g of water To able to have 74.86 g of HCl, the mass of solution 30.8 mass %HCl is = 74.86/30.8 * 100=243.05 g which contain : 243.05-74.86= 168.19 g of water We needed 319.14 g of water to produce desire product, so the water need to add in is 319.14-168.19= 150.95 g Final Answer: To make 394g of 19.0 mass % HCl, we add 150.95 g of water in 243.05 g of 30.8 mass %HCl
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