Write the following as liters only(remeber you will need to round off to the nearest liters.) a.3 546ml b.2 876ml c.9 234ml d.6 127 ml e.8 750ml f.9 500ml

104

1,000ml = 1 liter

We need common units to compare. We can convert everything to liters or milliliters, one or the other, it does not matter.

I will use liters.

30ml = 0.030 liters
We have 1.5 liters of dough.

How many times does 0.03 go into 1.5?

Good luck,

Paul

This is equal to about 61.6 ?L (U.S.) or 59.2 ?L (Britain). Pharmacists have since moved to metric measurements, with a drop being rounded to exactly 0.05 mL (50 ?L, that is, 20 drops per milliliter).

1 ?L is .000001 Liter so 1 million drops would equal one liter, US.

12.6802584595875 ounces

4000

A Milligram is MG and a Milliliter is a ML. (MG) is used in dry measure and (ML) is used in liquid measure

1000 micrograms (mcg) EQUALS 1 milligram (mg)
1000 milligrams (mg) EQUALS 1 gram (g)
1000 grams (g) EQUALS 1 kilogram (kg)

1 kilogram (kg) EQUALS 2.2046 pounds
1 ounce EQUALS 28.35 grams
1 pound EQUALS 0.4536 kilograms
1 pound EQUALS 16 ounces
16 ounces EQUALS 453.6 grams

1000 milliliters (mL) EQUALS 1 liter (L)
1 liter (L) EQUALS 1.0567 quarts liquid
1 quart liquid EQUALS 0.9463 liter
4 quarts EQUALS 1 gallon

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. !

1.678 ml

This problem can be easily solved by using proportions:
To use the proportions you have to be aware of the meaning of percent (%).
Percent aways refers to how much of something compared to 100 of it.

For example, as in your problem, there are 95 grams of Na2CO3 per 100 mL of solution.

Percent, when applied to solutions, is more commonly a mass per mass ratio times 100; but for this problem, since you did not specify, I am going to assume you meant grams per volume, since the problem is asking for the "amount" of the Na2CO3 solution, which seems to be the solution volume.

That is, the unknown for this problem is the volume of the 95% Na2CO3.

You want to prepare 5 liters (i.e., 5 L) of a 0.5% solution of Na2CO3.
You don't say what the unit is for the 0.5, but I am taking the liberty of assuming you meant 0.5%.

Another way to state what you want to prepare is this:
You want to prepare 5 L of Na2CO3 solution in which there are 0.5 grams of Na2CO3 per 100 mL of solution.

The percent ratio can be used as a conversion factor, which can be written as a ratio:
0.5 g Na2CO3 / 100 mL solution.
It is important to use the units (in boldface) for each numerical value, because they will guide you error-free in this type of problem! Remember, you can write it this way, because % in this problem means how much (in grams) per 100 (in mL).

****

OK, here is how you use this conversion factor: You multiply it by the given quantity, which is 5 L or 5,000 mL. (I converted the L to mL by multiplying it by 1000, since there are 1000 mL in 1 L - it's a good idea to memorize that.)
5000 mL x (0.5 g / 100 mL).
Remember, always multiply the given quantity (not a ratio) by the conversion factor (which is a ratio). You can assure accuracy by being sure that the like-units (mL) cancel out.
The calculated result of this math operation is 25 g (of the solute, Na2CO3).
Note that the unit for this result is g (or grams) of Na2CO3; you should always be sure to attach the complete unit to the numerical value 25. Also note that the calculated result is actually only a one-sigfig quantity; I indicated that by putting the sig fig in boldface and also underlining it.

****

Before you can complete this problem, you need to use one more conversion factor, already mentioned in the one-sentence 2nd paragraph above:
95 g Na2CO3 / 100 mL solution
However, you will have to invert or flip it as follows:
100 mL solution / 95 g Na2CO3
You will use it as above for the first conversion factor; you will multiply it by a different quantity (not a ratio), as follows:
25 g Na2CO3 x (100 mL solution / 95 g Na2CO3)
Do you see why I flipped the ratio now? To allow cancelling out of the like units, "g Na2CO3."

****

So, the answer to this problem is 26.32 mL of the 95% Na2CO3 solution (before proper rounding off).

After rounding off using the rules for handling sig figs,
the correct answer is 30 mL of the 95% solution of Na2CO3.

Additional Notes
Some information about rounding off
Notice that in the above paragraph, and in the two paragraphs before that, I have underlined the only digit that is considered to be a significant digit (sig fig). An important rule to remember regarding sig figs and rounding off in math operations involving muliplication or division is this:
The calculated result may not have more sig figs in it than was in the quantity with the smallest number of sig figs. So there can be only one sig fig in 26.32, because it was calculated using another one-sigfig number, 25. (The 25 was obtained from calculation using another one-sigfig number, 0.5). To learn more about sig figs and how to consider them properly when rounding off, you may consult almost any chemistry textbook, or Google up on the world wide web.

Regarding the question implied by the problem, it is not based on a realistic situation. The reason I say this is because there is no such thing as a 95% aqueous solution of Na2CO3. Only about 25 grams will dissolve in 100 mL of water. (This fact may be confirmed by looking up the solubility of Na2CO3 in water.)

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The bike will actually be running leaner at 32:1, than at 40:1.

You need to put 250 ml to 300 ml

Regards