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Q: How would a solution that is labeled 5.0 M be read as?

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50 g

Let's say the total solution is 100 liters. 50 of the liters is glucose and 50 is water. We want to make the 50 glucose equal to 10% of the total solution. For that to happen, we need to make the total solution 500 liters (50 of the 500 would be a 10% solution). So we add 400 liters of water to the original 100 liter (50/50) solution. Take the total number of units and multiply by 4. Add that much in water.

50

You can use a graduated cylinder or a pipette of 50 mL.

The volume reach the half.

A 50 ml solution that is 10% acid will consist of 5 ml of acid (10% of the volume) and 45 ml of water (90% of the volume). You're not adding any water, but you want to add enough acid to make a solution that is 50% acid and 50% water. You will need to have a total of 45 ml of acid in the mixture to make it a 50/50 solution, since the amount of water is also 45 ml. You have 5 ml in there already, so you would need to add 40 ml of acid. That would make a total 90 ml solution that is 50% water (45 ml) and 50% acid (45 ml).

How much 50 percent antifreeze solution and 40 percent antifreeze solution should be combined to give 50 gallons of 46 percent antifreeze solution?

first calculate vd * CD =Vc * Cc Vc=50 * 2 / 5 =20ml

A 38 out of 50 is a 76%. This is calculated by dividing 38 by 50 (38/50). You then multiply the result by 100, and the solution is your percentage out of 100.

In order to reduce the percentage of acid from 80% to 50%, you would need to add another 36 kg of diluent (e.g. water).

50 percent of the dissolved minerals will condense out of solution by crystallization.

100 g of solution containing 50 g of NaOH.