84 mg/ml, or 1680 mg/20 ml
The amount of salt that can dissolve in 20mL of water depends on the solubility of the salt at that temperature. For common table salt (sodium chloride), approximately 36 grams can dissolve in 20mL of water at room temperature.
To neutralize the sulfuric acid completely, you need a 1:2 molar ratio of sodium hydroxide to sulfuric acid. Therefore, you would need to add twice the amount of sodium hydroxide compared to the amount of sulfuric acid, which is 40.0 mL of the sodium hydroxide solution.
Yes, a solution is considered concentrated when there is a large amount of solute (salt in this case) dissolved in a given amount of solvent (water in this case). With 20ml of salt in 50ml of water, the concentration of the solution would likely be high.
The formula for this question is c1v1= c2v2 c1v1= 20ml x 0.01m = 0.02 x 0.01 c2v2= 0.03c2 therefore it is (0.02 x 0.01) / 0.03 = c2 c2 is the molarity of our na-oh solution cheers :D The formula for this question is c1v1= c2v2 c1v1= 20ml x 0.01m = 0.02 x 0.01 c2v2= 0.03c2 therefore it is (0.02 x 0.01) / 0.03 = c2 c2 is the molarity of our na-oh solution.
To determine the number of moles of lithium in 20mL of 0.5mM lithium phosphate, first calculate the moles of lithium in 1mL of the solution: 0.5mM = 0.5 mmol/L = 0.5 × 10^-3 mol/L = 5 × 10^-4 mol/L Now, for 20mL: 5 × 10^-4 mol/L × 20 mL = 0.01 moles of lithium
20ml
20mL
20ml / 100 * 0.5 = 0.1ml
The amount of salt that can dissolve in 20mL of water depends on the solubility of the salt at that temperature. For common table salt (sodium chloride), approximately 36 grams can dissolve in 20mL of water at room temperature.
20ml
A. 2 g
20ml - 0.2ml = 19.8
The rise in temperature is affected by the amount of anhydrous sodium carbohydrate added to 20ml of water
20mL equates to about 4.1 (4.05768) US teaspoons.
5ml = 1 teaspoon 20ml = 4 teaspoons
There are 3 units of 20ml in 60ml, because 20ml x 3 = 60ml.
1 tablespoon