How many milliliters of sulfuric acid do I need to make 98 sulfuric acid to 12 sulfuric acid? How ma

How many milliliters of sulfuric acid do I need to make 98 sulfuric acid to 12 sulfuric acid? How many milliliters of waterAnswer: Let the concentrated sulfuric acid be taken x milliliters, and the water y grams have the following algebraic formula:

x∶y===1∶

In other words, when taking one milliliter of concentrated sulfuric acid, you must take 13.19 grams of water, and so on, when you take ten milliliters of concentrated sulfuric acid, you must take 131.9 grams of water to make 12% dilute sulfuric acid. Because it is known that the density of 98% concentrated sulfuric acid is slightly in units).

The density of sulfuric acid with a mass fraction of 98% is, so the mass of a milliliter of concentrated sulfuric acid is grams.

Set to add x ml of water (density of 1 g ml) with a mass of x grams.

So, there is: solution x=989a 75 grams.

That is, to dilute a milliliter of 98% concentrated sulfuric acid into 12% dilute sulfuric acid, York water is required, in milliliters.

98% sulfuric acid density, water density 1.

Cross-type of precedent quality.

The mass ratio of concentrated sulfuric acid to water is 12:86, divided by density, and the volume ratio is 76:1000

Therefore, if 1000 ml of water is used for the configuration of 12% sulfuric acid, then 76 ml of concentrated sulfuric acid is required.

It is calculated by using the principle that the mass of the substance does not change before and after dilution of sulfuric acid.

Original concentration * required volume = dilute concentration * dilute volume.

Categories: Education, Science, >> Science & Technology.

Analysis: With this family to make 100 grams of 10% sulfuric acid, 98% of the sulfuric acid needs to be diluted, and the quality of sulfuric acid of the diluted dissolution material remains unchanged.

m dilute * p dilute % = m concentrated * p concentrated %

100g*10%=m-98%.

m concentration = g m water = 100 g g = g.

Because 98% concentrated sulfuric acid has a density of gram cm 3

So v concentration = m concentration = g (gram dust liquid cm 3) = ml.

v water = m water water = g (1 g cm 3) = ml.

Summary. To prepare 100 grams of 40% sulfuric acid, 98% concentrated sulfuric acid and 18% dilute sulfuric acid are required.

Can you hurry.

I'm in a bit of a hurry here.

Good. Thank you.

Thank you. You're welcome! This problem is easier to understand with equations, the main reason is that the result is not an integer, so the calculation is a little more troublesome.

The mass of sulfuric acid contained in 100g of 10% dilute sulfuric acid is: 100 * 98% concentrated sulfuric acid is required: 10 (

The required water is: volume is:

Solution: The mass of 10% sulfuric acid solution can be prepared as x

Depending on the quality of the solute before and after dilution, it does not become:

10ml*x=mass of water to be added=

Then the volume of water v=m p= is required

Let the mass of concentrated sulfuric acid of 10 be xg, then 10 1·84 98 = x = of v 1 10 solution

The required water v= The question is very simple, study hard, classmates.

It should be milliliters, let the amount of water be x, and the equation is 10ml multiplied by x and then multiplied by 10% equals 98% multiplied by 10.

The density of sulfuric acid with a mass fraction of 98% is, so the mass of a milliliter of concentrated sulfuric acid is grams.

Set to add x ml of water (density of 1 g ml) with a mass of x grams.

So, there is: solution x=989a 75 grams.

That is, diluting a milliliter of 98% concentrated sulfuric acid into 12% dilute sulfuric acid, you need York water, containing Shihe milliliters. Talk about Zheng Limb.

98% sulfuric acid density, water density 1.

Cross-type of precedent quality.

The mass ratio of concentrated sulfuric acid to water is 12:86, divided by the density, and the volume ratio is 76:1000

Therefore, if you use 1000 ml of water with 12% sulfuric acid, then you need 76 ml of concentrated sulfuric acid.

To prepare 100 grams of 10% sulfuric acid, 98% of the sulfuric acid needs to be diluted, and the quality of the dilute solute sulfuric acid remains unchanged.

m dilute * p dilute % = m concentrated * p concentrated %

100g*10%=m-98%.

m concentration = grams. m water = 100 g g = g.

Because 98% concentrated sulfuric acid has a density of gram cm 3

So v thick = m thick thick = g (g trillion cm 3) = milli cherry guess to do the rise of the ridge balance.

v water = m water water = g (1 g cm 3) = ml.

It is calculated by using the principle that the mass of the substance does not change before and after dilution of sulfuric acid.

Original concentration * required volume = dilute concentration * dilute volume.

98% sulfuric acid specific gravity = g ml, 20% sulfuric acid specific gravity = g ml, 10% sulfuric acid specific gravity = g ml (1) 500 ml of 20% sulfuric acid, weight = 500 * g), containing pure sulfuric acid = g).

Converted into 98% sulfuric acid, weight = grams), volume = ml) so it is good to be bored: to prepare 500 ml of 20% sulfuric acid, 98% of sulfuric acid is required, water (ml) (2) 5000 ml of 10% sulfuric acid, weight = 5000 * g), containing pure sulfuric acid = 5330 * 10% = 533 (g).

Converted into 98% sulfuric acid, weight = 533 98% = gramming lead), volume = ml) Therefore: to prepare 5000 ml of 10% sulfuric acid, 98% sulfuric acid is required in milliliters of water (ml).

Here, during the dilution of sulfuric acid, the specific gravity of the solution changes, and therefore the volume changes, and what does not change is the weight. Therefore, the amount of water is calculated based on the conservation of weight.

Due to the discrepancies in the specific gravity data of sulfuric acid cited, the calculations may differ slightly.

3) Zinc sulfate is solid. Generally, it is a crystal containing seven crystals of water. Zinc sulfate is easily soluble in water to form an aqueous solution.

Sulfuric acid = 200 98% 10% = 1960 grams.

Add water = 1960-200 = 1760 grams.

If you don't understand, you can ask, it helps, thank you!