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How much 5 percent acid must be added to 25 percent acid to get 200 ml of 8 percent acid?
Answer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solutionLet x = 5% acid solutiony = 25% acid solution Equations:x + y = 200mL ----equation (1)0.05x + 0.25y = 0.08 * 2000.05x + 0.25y = 16multiplying by 1005x + 25y = 1600 ----equation (2)eliminating equations (1) and (2)-5(x + y = 200)-5-5x -5y = -10005x +25y = 1600=========20y = 600y=30substitute x=30 to equation (1)x + 30 = 200x = 200 - 30x = 170thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution
What is bigger 200ml or 2l?
2 liters (2L) is significantly larger than 200 milliliters (200ml). In fact, 2 liters is equivalent to 2000 milliliters, making it ten times the volume of 200ml. Therefore, 2L is the larger measurement.
How many is 200ml?
That is 7.055 ounces.
What is 200ml of 1 l?
66.67
How many 200ml bottles cordial can filled from a3l jug?
You can fill 15 bottles of 200ml with a 3L jug of cordial.
Show by calculation how would you prepare 200ml of 80 percent glycerol?
To prepare 200ml of 80% glycerol solution, you would mix 160ml of glycerol (200ml x 80%) with 40ml of water. Calculations: 200ml x 80% = 160ml glycerol, and 200ml - 160ml = 40ml water needed. Mix the glycerol and water thoroughly to obtain the 80% glycerol solution.
Which volume is smaller 150mL 11mL 200mL?
11 mL is the smallest volume of the three.
How do you prepare 200ml of ethyl acetate of 0.1 Normality solution?
To prepare 200ml of 0.1 N ethyl acetate solution, you will need to calculate the amount of ethyl acetate needed. Since the molecular weight of ethyl acetate is around 88.11 g/mol, for 200ml of 0.1 N solution, you would need around 1.76g of ethyl acetate. Dissolve this amount of ethyl acetate in distilled water to make up the final volume to 200ml.
What is 10 percent of 200ml?
20ml
When 200 mL of water are added to 100mL of 12 percent KCL solution the final concentration of KCL is?
The initial amount of KCl in the solution is 12mL (12% of 100mL). When 200mL of water is added, the total volume becomes 300mL. You would then divide the initial amount of KCl (12mL) by the total volume (300mL) and multiply by 100 to get the final concentration of KCl in the solution.
You are provided with a 1000ppm sodium solution describe how you would prepare a volume of 200ml with a concentration of 5ppm sodium in a two step dilution?
To prepare a 200ml solution with a sodium concentration of 5ppm from a 1000ppm stock solution, you will first dilute the stock solution by a factor of 200. So, take 0.2ml of the 1000ppm solution and dilute it to 40ml with water. This will give you a solution with a concentration of 50ppm. Then, take 0.4ml of this 50ppm solution and dilute it to a final volume of 200ml with water to achieve the desired 5ppm sodium concentration.
What is the molarity of a solution when 50.0ml of a 4.00M KOH solution is diluted to 200ml. what is the percent concentration of KOH in this final solution. asume the density of 1.0gmole?
The concentration is 1 mol/L or 5,611 g KOH/100 mL solution.
What is the density of a liquid that has a mass number of 200 g and a volume of 200ml?
d=m/v so 200g/200ml= 1g/ml
How much 5 percent acid must be added to 25 percent acid to get 200 ml of 8 percent acid?
Answer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solutionLet x = 5% acid solutiony = 25% acid solution Equations:x + y = 200mL ----equation (1)0.05x + 0.25y = 0.08 * 2000.05x + 0.25y = 16multiplying by 1005x + 25y = 1600 ----equation (2)eliminating equations (1) and (2)-5(x + y = 200)-5-5x -5y = -10005x +25y = 1600=========20y = 600y=30substitute x=30 to equation (1)x + 30 = 200x = 200 - 30x = 170thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution
What is the density of an object that has a mass is 500 grams and a volume of 200ml?
2.5
What is the total volume of 200mg in 200ml?
The volume of 200 ml is 200 millilitres. 200 milligrams is irrelevant to the question.
How many equivalent of H2SO4 are contained in 200ml of a 0.2 N solution?
a lot
