One percent by weight.
Alcohol has a density of less than water, so 1 g of alcohol is more than 1 mL.
Suppose there is x% by volume of something in a mixture then it simply means out of 100 ml of that sample x ml is of that thing. ------------------------------------------------------------------- There are two common ways of indicating how much of each element or compound is contained in a mixture. They are percent by volume and percent by weight. For example, the amount of alcohol in an alcoholic beverage is measured in proof, where one proof equals one half percent alcohol by volume.
Percent of an objects mass is expressed in terms of its weight. Percent of an objects volume is expressed in terms of its size.
If the percents given are by weight or mass, this is very straightforward: The ratio between the desired percentage and the initial percentage is 1/50. Therefore, a given mass of initial solution must be diluted to 50 times its original mass to obtain the desired lower concentration, or in other words, 49 parts of diluent must be mixed with each part of initial solution. If the percents involve volume measurements, it would be necessary to take into account and change in density occasioned by the dilution.
That depends on the volume of the brick. Whatever its volume is, its weight underwater is(weight of the brick in air) minus (weight of an equal volume of water)
Bulk density = dry weight / volume, then by knowing the dry weight and bulk density we can calculate the volume.
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
A 1% solution normally contains 1 gram of active ingredient per 100 ml of solution (weight-volume percent) Could also be 1gm per 100 gms (weight-weight percent)- but normally weight-volume is used.
It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.
Calculate the weight of sucrose for the desired volume and concentration of the solution.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
The concentration of the diluted solution will be 15(300/1000) = 4.5 %, if the percent is expressed on a weight/volume basis.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
% volume
% volume
Suppose there is x% by volume of something in a mixture then it simply means out of 100 ml of that sample x ml is of that thing. ------------------------------------------------------------------- There are two common ways of indicating how much of each element or compound is contained in a mixture. They are percent by volume and percent by weight. For example, the amount of alcohol in an alcoholic beverage is measured in proof, where one proof equals one half percent alcohol by volume.
Mole percent vs volume percent is a simple comparison to make. Mole percent is equivalent to moles of substance/moles of total x 100. Volume percent is volume of substance/volume of total x 100.