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If you have a 70 percent solution of dextrose How many grams of dextrose is in 400ml of this solution?

In a 70% dextrose solution, 70% of the total weight is dextrose. To calculate the grams of dextrose in 400ml of this solution, you would multiply 400ml by 70% (or 0.70) to find the amount of dextrose present.


What do you expect to observe when a solution of potassium nitrate saturated at 343 k or 70 c is cooled to room temprature?

Upon cooling the solution of potassium nitrate from 70°C to room temperature, you would expect to observe the formation of crystals as the solubility of potassium nitrate decreases with decreasing temperature. These crystals will form as the excess potassium nitrate in the solution starts to come out of the solution and solidify.


What is the solvent in 70 percent alcohol?

70% alcohol signifies the amount the alcohol present in 100 ml of the solution. Considering the solvent to be water, 70% alcohol contains 70ml alcohol and 30ml water. Thus, the total comes out to 70 + 30 = 100ml. That's what the "%" signifies - (out of 100!)


Why 70 percent alcohol is used for sterilization?

70% alcohol solution is more effective for sterilization compared to higher concentrations like 100% alcohol because it has a lower evaporation rate, allowing it to stay in contact with the bacteria or viruses longer to kill them. A 70% alcohol solution also penetrates the bacterial cell membrane more effectively, denaturing proteins and disrupting cell function. Additionally, 70% alcohol is less likely to damage surfaces or instruments being sterilized.


How do you prepare 5 percent nitric acid solution from 55 percent nitric acid?

To solve this problem, we basically have 2 equations and 2 unknowns. The unknowns are the (volume of water) & the (volume of 70 wt%) nitric acid to add. * This problem will assume that you are interested in making 1 L (or 1000 mL) of 5 wt% nitric acid solution. Equation 1: (volume of water) + (volume of 70 wt% nitric acid) = 1000 mL Equation 2: mass of nitric acid / [mass of water + mass of 70 wt% nitric acid solution] = 0.05 (0.05 is 5 wt%) * Remember that mass = density * volume * Remember that 70 wt% nitric acid solution mean that for 100 grams (gm) of this acid, then there's 70 grams of HNO3 * Remember that density of 70 wt% nitric acid solution is 1.413 gm/cm^3 * Remember that density of water is 1 gm/cm^3 Equation 2 is now re-written as: [(density of 70 wt% nitric acid soln)*(volume of 70 wt% nitric acid)*0.70] / [(volume water)*(1gm*cm^3) + (volume of 70 wt% nitric acid)*(1.413gm/cm^3)] = 0.05 Solving for the 2 equations gives answer to the 2 unknowns: Answer: To make 1000 mL of 5 wt% nitric acid solution, add 1) 51.63 mL of 70 wt% nitric acid solution 2) 948.37 mL of water