#### Application Information

This drug has been submitted to the FDA under the reference 013157/002.

#### Names and composition

"ANHYDRON" is the commercial name of a drug composed of CYCLOTHIAZIDE.

#### Forms

ApplId/ProductId Drug name Active ingredient Form Strenght
013157/002 ANHYDRON CYCLOTHIAZIDE TABLET/ORAL 2MG

#### Similar Active Ingredient

ApplId/ProductId Drug name Active ingredient Form Strenght
013157/002 ANHYDRON CYCLOTHIAZIDE TABLET/ORAL 2MG
018173/001 FLUIDIL CYCLOTHIAZIDE TABLET/ORAL 2MG

#### Ask a doctor

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##### The molecular weight of anhydrons calcium chloride is 110.99. How many grams.....?
How many grams do you need for 100ml of 5mM calcium chloride? Asked by Oswaldo Zamarron 2 years ago.

You have the volume and molarity of CaCl2 required. You need to first calculate the number of moles of calcium chloride. No. of moles of CaCl2 required = 0.1L x 5mM (or 5 mmol/L) = 0.5 mmol (or 0.0005 mol) Given the molecular weight, the amount of anhydrous calcium chloride to weigh out is thus: 110.99 g/mol x 0.0005mol = 0.55495g approx. 0.555g ========================= No. of moles (mol) = vol (L) x concentration (M) Mass (g) = relative atomic/molecular weight (g/mol) x no. of moles (mol) **be careful when converting units like mmol to mol, and mL to L, etc.** Answered by Tiffaney Gipe 2 years ago.

Calcium Chloride Molecular Weight Answered by Hellen Sappenfield 2 years ago.

Calcium Molecular Weight Answered by Judie Quirke 2 years ago.

This Site Might Help You. RE: The molecular weight of anhydrons calcium chloride is 110.99. How many grams.....? How many grams do you need for 100ml of 5mM calcium chloride? Answered by Talisha Silvester 2 years ago.

Molarity = mole of solute / Litre of solvent mM = 1000mole/L 5mM = 0.005mol/L volume of solvent = 100 ml /1000= 0.1 L mole of calcium chloride = 0.005 mol /L * 0.1 L =0.0005 mol weight of calcium chloride = mole * molecular weight = 0.0005* 110.99 = 0.055 g Answered by Clarine Weisser 2 years ago.

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##### Need chemistry help?
I have a lab that I dont know how to do the calculationsmy objective is to determine the formula of the hydrate of Magnesium Sulfate (MgSo4)so how do i get the ratio of MgSo4 to H2O here's my info mass of empty crucible =9.675 gramsmass of hydrate= 2.512 gramsmass of anhydrons MgSo4= 10.950 grams... Asked by Tisha Danis 2 years ago.

I have a lab that I dont know how to do the calculations my objective is to determine the formula of the hydrate of Magnesium Sulfate (MgSo4) so how do i get the ratio of MgSo4 to H2O here's my info mass of empty crucible =9.675 grams mass of hydrate= 2.512 grams mass of anhydrons MgSo4= 10.950 grams *mass of hydrate and mass of anhydrons will the mass of H2O so how do i get the ratio Answered by Dorinda Ardizzone 2 years ago.

10.950 g - 9.675 g = mass of anhydrous MgSO4 1.275 g = MgSO4 2.512 g - 1.275g = 1.237 g H2O 1.237 g H2O / 18.016 g/mol = 0.0687 mol H2O 1.275 g MgSO4 / 120.386 g/mol = 0.0106 mol MgSO4 Divide Moles of each by the smallest 0.0687 / 0.0106 = 6.48 0.0106 / 0.0106 = 1 You experimental data yields MgSO4 . 6.5H2O I would guess that you failed to drive off all of the water. I believe the correct value would be a 1:7 ratio- MgSO4.7H2O Answered by Grisel Clovis 2 years ago.

It is not clear to me what numbers you have. mass of empty crucible =9.675 grams mass of hydrate= 2.512 grams mass of anhydrons MgSo4= 10.950 grams - is this the mass of the anhydrous PLUS THE CRUCIBLE, ie, the anhydrous mass is 1.275 grams ? If so, you have 1.275 g MgSO4 picking up 1.237 g H2O. Or losing, if you started with the hydrate and calcined it, MW MgSO4 = 24.32 + 32.06 + 4(16) = 120.38, so you started with .0106 gmols MgSO4 and it picked up .0687 gmol H2) .0687/.0106 = 6.48 Epsom salts is the heptahydrate, MgSO4*7H2O, which I suspect is what you got. if a mere 8% of the anhydrous did not react, that would give the 6.48 I calculated [or if the hydrate did not fully calcine to the anhydride]. There is also MgSO4*H2O, the monohydrate. If you had a mixture of .913 heptahydrate and .087 monohydrate that would also give the 6.48 above. I do not see any other hydrates listed in my 1949 CRC handbook [admittedly rather old, but I doubt much changed as to hydrates of MgSO4]. If there were other hydrates, you could have formed other mixtures of various hydrates. To make the heptahydrate only, you would have to have picked up [or lost] 1.334 g H20, a difference of .0973 grams, with a hydrate weight of 1.372 g, with crucible 12.322 g Answered by Seymour Fay 2 years ago.

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