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1.3g
Molarity = moles of solute(CuSO4)/volume of solution(Liters) 0.967 grams CuSO4 (1 mole CuSO4/159.62 grams) = 0.00606 moles CuSO4 Molarity = 0.00606 moles/0.020 liters = 0.303 Molarity
The molarity is 6.
In titrations, the end point is when you have brought the tested sample to absolute neutral. At this point, if you add one more drop of titrating solution to the sample, you would change the pH sufficient to change the color of the indicator in the sample. This is the point at which you can determine the pH of the original solution, by calculating back the amount of titrating solution you had to add to the sample to neutralize it.
You can prove there is glucose in a sample by using Benedict's Solution. Heat up the sample, and add the Benedict's Solution. Assuming the solution is clear, if glucose is present it will change colour to red, or yellow, or green. If not, it will stay clear.
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What is the molarity of an HCl solution if 43.6 mL of a 0.125 M KOH solution are needed to titrate a 25.0 mL sample of the acid according to the equation below?
22
1.3g
1.0 Moles
Molarity = moles of solute(CuSO4)/volume of solution(Liters) 0.967 grams CuSO4 (1 mole CuSO4/159.62 grams) = 0.00606 moles CuSO4 Molarity = 0.00606 moles/0.020 liters = 0.303 Molarity
The molarity is 6.
The question, as worded, is a little ambiguous. Rather, the question you should be asking is “What is the molarity of a 125 ml aqueous solution containing 10.0g of acetone?” Acetone is roughly 58 grams per mole. Therefore, a 125 mil solution with 10 g of acetone would contain roughly 0.17 moles, and the molarity would be roughly 1.4See the Related Questions for more information about how to calculate the molarity of a solution
ammonium oxalate is added to calcium carbonate because in the reaction between the two a crystal is formed that contain the Ca+2 ion. This is useful because if you have a sample of sodium carbonate with an unknown molarity you can use the oxalate to extract this calcium and determine what the molarity of the unknown solution was
first measure the volume of the sample solution needed to change the blue color of the DCPIP solution into colourless. then, weigh the mass of the sample solution. finally calculate the concentration by using the formula: volume required t change the color of DCPIP solution (dm) per mass of the sample solution (g)
In titrations, the end point is when you have brought the tested sample to absolute neutral. At this point, if you add one more drop of titrating solution to the sample, you would change the pH sufficient to change the color of the indicator in the sample. This is the point at which you can determine the pH of the original solution, by calculating back the amount of titrating solution you had to add to the sample to neutralize it.
The molarity is 0.001255. Should you really be asking an AP Chem question on Wiki Answers, anyways?