4.00 mL of a 2.00 x 10 to the -3 M solution of NaSCN is diluted to 100 mL of solution. What is the molarity of SCN to the minus solution?

What is the original concentration of your 100 ml?

To dilute 100ml of something to 2.5% by volume, first calculate 100 divided by the desired % 100/2.5 = 40.
So you would want to end up with 40 times the volume, or 4000 ml (4L). Add 3900 ml water (3.9L) because 100ml +3900ml = 4000 ml
https://en.wikipedia.org/wiki/Dilution_(equation)

But that assumes that your 100 ml is pure, or 100% concentration of your active ingredient. If your 100ml is already diluted to a certain percent, you have to reduce the added water in proportion.

Try this...
https://answers.yahoo.com/question/index?qid=20090713132509AAD6ryn

1.0 LITER = 1000 ml so 100 ml = 0.10 Liter

or one tenth so you need to divide 58.44 by 10 with the result being

TA DA 5.844 gms (be sure you read the decimal points correctly. !

You cannot convert a unit of mass (mg) to a unit of capacity (volume). Mass and volume of a specific substance are related through the mass density (mass per unit volume).
What you have seems to be a dilution problem but it is not formulated completely. Please try again with necessary details.

I need to be sure what you meant by the m symbol: M or m. The first is used for molarity and the latter for molality. If it is molality we do need to know the masses per unit volume for the solutions (the so-called densities)

However if you meant 2.5 M and 0.3 M, then the problem is readily solved. I 'll take a step by step method.
2.50 M means that the original solution contains 2.5 mol of solute per 1 L of solution. Let x be the volume of the solution.
The amount of solute ( expressed as a number of moles) isAmount of solute in original solution (in mol)=2.5 (mol/L)*x (L)
Now consider the final solution.
The amount of solute in the final solution is given by:Amount of solute (in mol) =0.3 mol/L*2 L=0.6 mol
However, this last value is exactly the amount of solute the original solution, because you are just putting it in a bigger volume (dilution)
Equating the two amounts of solute we get
2.5 x=0.6 (in mol) and
x=0.6/2,5=0.24 L=240 mL
When concentrations are given in molar units, the foregoing reasoning can be summarized as
initial concentration (c_i) * initial volume (V_i)=final concentration (c_f)*final volume (V_f) or more simply
c_i*V_i=c_f*V_f
Using the formula to find the initial volume
V_i=(c_f*V_f)/c_i=(0.3 *2)/2.5=0.24
The starting volume is 0.24 L or 240 mL

I am going to assume your term, "nf," which I do not recognize as an acceptable unit of the metric system, is meant to be molarity (M). So, it seems to me that you are asking for the molarity of the solution that is indicated in the problem.

To calculate the molarity of the solution of NaOH in this problem, one must apply the ratio, moles per liter of solution, the definition of molarity (moles/L).

A 27.5% solution means (27.5 grams) / (100 grams of solution), which in this case is equivalent to
27.5 g / 100 mL solution, because the density of water is 1.00 g/mL.

Since 100 g solution is specified, that means there are 27.5 g of NaOH in the solution (100 mL volume, or 0.100 L).

Moles of NaOH in solution = [27.5 g NaOH / 40.0 g mol NaOH/ mol NaOH] = 0.688 mol NaOH.

Therefore, molarity of NaOH = mol NaOH / L = (0.688 mol NaOH / 0.100 L) = 6.88 M NaOH.

###

Hi,

In chemistry , concentration is the measure of how much of a given substance there ... solution \right )%Molarity M\ ... m**)Molar ... or F)Normality N\ ...

Hope it will help you.

100 ml of 2cycle oil to 4000 ml of fuel..Which is 40/1 ratio
100 ml of 2cycle oil to 5000 ml of fuel.. which is 50/1 ratio

4 o/z is 118.29411875006815 ml to 1 gallon of fuel (3.79 litres or 3790 ml) Which means 32/1 ratio...I find that to confusing,so it is easier to say 100 ml of oil to 3200 ml of fuel ===32/1.

Just bought a P120 from craigslist. A computer recycle company was selling it as parts for $150. They said it has the SCN error code and the keys don't play. It played the demos only.. Payed $100; took it home, opened it up & found the keyboard plug had detached from the motherboard. Plugged it in & BAM! It works just fine.

Hello, I can help you if you include more data in your question.

As an organic compound, lidocaine is insoluble in water, so I am assuming that either some combination of water and alcohol (ethanol), or a pure organic solvent like ethanol or diethyl ether is being used.

If lidocaine were soluble in water (which it's not), you would have been able to solve your problem as indicated below. Maybe you would be able to modify the indicated solution by taking into consideration the density of the organic solvent.

Percentage (%) by mass is one of the common ways to express the concentrations of solutions.
Since % means the number of grams of solute in 100 grams of solution (solute + solvent), a 1.5% solution of any solute refers to 1.5 grams of it in 100 grams of solution.

Assuming we are referring to an aqueous solution (that is, where water is the solvent), this reasonably approximates to 1.5 grams of lidocaine per 100 mL solution. We can assume this result, because we have a dilute watery solution, which probably has the density close to the density of pure water, which is 1.0 gram per 1.0 mL at commonly used temperatures, such as average room temperature.

So, 1.5 g lidocaine per 100 mL solution can easily be scaled down by one-tenth to give:
0.15 g lidocaine per 10 mL solution.

###