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How much gram of fecl3 to be used in preparing fecl3 solution for testing citric acid in semen?
The amount of FeCl3 needed depends on the concentration of the FeCl3 solution required for the test. Typically, a 2-5% solution of FeCl3 is used. To make a 100mL of 2-5% FeCl3 solution, you would need to dissolve 2-5 grams of FeCl3 in distilled water. The exact amount can be calculated using the formula: (desired % concentration/100) x volume of solution needed x molar mass of FeCl3.
What is the calculation for 0.1 N naoh solution?
To calculate the concentration of a 0.1 N NaOH solution, you need to know the molar mass of NaOH which is approximately 40 g/mol. Since 1 N solution is equivalent to 1 gram equivalent weight of the solute dissolved in 1 liter of the solution, a 0.1 N NaOH solution would contain 4 g of NaOH per liter of solution.
How do you prepare 0.5N NaOh?
Take half volume of 1.0 M NaOH and add another half volume of water. Or Take 20.0 gram NaOH , carefully dissolve this completely in ca. 0.9 L water and then fill up to 1.0 L end volume.
How will you prepare 1 percent naoh?
To prepare a 1% NaOH solution, you can mix 1 part of NaOH (sodium hydroxide) with 99 parts of water by weight. For example, to prepare 100 mL of 1% NaOH solution, you would dissolve 1 gram of NaOH in 99 grams of water. Remember to always add NaOH to water slowly while stirring to avoid splattering and heat generation.
How many grams does 2 moles of NaOH weighs?
to solve this you must have the molecular weight of NaoH and you can get this by add the molar masses together (H=1 , O=16 , Na-23 ) then multiply the no of moles by the molecular mass that is for mole gram conversion so you have (23+16+1)= 40 and already you have multiplied (2.40)=80 then 2 . 40 = 120 then 80 / 40 = 2 the answer is 2
How many moles are in gram of NaOH?
moles = mass/Mr moles = 100/(23+16+1) moles of NaOH = 2.5mol
How much gram of fecl3 to be used in preparing fecl3 solution for testing citric acid in semen?
The amount of FeCl3 needed depends on the concentration of the FeCl3 solution required for the test. Typically, a 2-5% solution of FeCl3 is used. To make a 100mL of 2-5% FeCl3 solution, you would need to dissolve 2-5 grams of FeCl3 in distilled water. The exact amount can be calculated using the formula: (desired % concentration/100) x volume of solution needed x molar mass of FeCl3.
What is the percent weight of 100 grams FECl3 and 900 mL of water?
Many people think that the gram is a measurement of volume. However, the gram is a measurement of weight, and so 100 grams weighs 100 grams or 3.52739619 ounces.Water weighs 8.34 (rounded to the nearest hundredth) pounds. 900 millilitres is equal to 0.237754846 US gallons. Therefore, 900 millilitres of water weighs about 1.98287542 pounds or 899.417161 grams.
What is the calculation for 0.1 N naoh solution?
To calculate the concentration of a 0.1 N NaOH solution, you need to know the molar mass of NaOH which is approximately 40 g/mol. Since 1 N solution is equivalent to 1 gram equivalent weight of the solute dissolved in 1 liter of the solution, a 0.1 N NaOH solution would contain 4 g of NaOH per liter of solution.
How do you prepare 0.5N NaOh?
Take half volume of 1.0 M NaOH and add another half volume of water. Or Take 20.0 gram NaOH , carefully dissolve this completely in ca. 0.9 L water and then fill up to 1.0 L end volume.
Why is chlamydia gram negative?
it explains its added resistance. it is gram negative due to it's enhanced cell wall, etc.
How will you prepare 1 percent naoh?
To prepare a 1% NaOH solution, you can mix 1 part of NaOH (sodium hydroxide) with 99 parts of water by weight. For example, to prepare 100 mL of 1% NaOH solution, you would dissolve 1 gram of NaOH in 99 grams of water. Remember to always add NaOH to water slowly while stirring to avoid splattering and heat generation.
If 1.20 grams of sodium metal are reacted with 3.00 ml of water how many grams of NaOH will be produced?
First write balance equation 2Na + 2H2O --> 2NaOH + H2 molar mass Na = 22.99 g/mol molar mass H2O = 18.02 g/mol molar mass NaOH = 40.0 g/mol Determine limiting reagent. 1.20 g Na * 1mol Na/22.99g Na = 0.05219 (0.0522) mol Na Since 2Na = 2NaOH in balanced equation, The mol of NaOH is also 0.0522 3ml h20 = 3 gram h20 (1ml^3=1g^3) 3g H20 * 1mol/18gH20 = 0.167 (0.17) mol H20 So .17 ml H20 equals .17 mol NaOH and .0522 mole Na equals .0522 mol NaOH The smaller one is the limiting reagent, which in this case is Na 0.0522mol NaOH*40g NaOH/1mol NaOH = 2.09 gram NaOH
How gram positive bacilli defer from gram negative bacilli in gram stain?
First gentian violet or crystal violet solutn is applied,they bcome purple now iodine is added and acetone or ethyl alcohol is applied gram -ve bacteria lose their color while gram positive bacteria don't
The latent heat of vaporization of ethanol is 200 calories gram. This means that?
about 200 calories of heat must be added to 1 gram of ethanol to convert it from a liquid to a gas
How many grams does 2 moles of NaOH weighs?
to solve this you must have the molecular weight of NaoH and you can get this by add the molar masses together (H=1 , O=16 , Na-23 ) then multiply the no of moles by the molecular mass that is for mole gram conversion so you have (23+16+1)= 40 and already you have multiplied (2.40)=80 then 2 . 40 = 120 then 80 / 40 = 2 the answer is 2
How do you prepare 0.01 N NaOH from 0.02 N NaOH?
Since "normality" is defined as the gram equivalent weight of a substance in a liter of solution, a 0.02 N NaOH solution would have 0.02 gram equivalents of NaOH per liter. To reduce it to 0.01 N you need only dilute it to one half of the original - e.g. 500 ml of NaOH mixed with 500 ml of pure water. Because there is a small change in the density upon mixing, the exact amount of water will differ slightly from 500 ml, but for a solution as dilute as 0.02 N, it won't be that far off. The best way to get it exact would be to start with a known volume of 0.02 N NaOH and then add enough water to bring the total to exactly twice the original volume. This might be accomplished by doing it in a graduated cylinder or adding it from a burette into a volumetric flask. the important thing is to know the starting volume of the 0.02 N solution and the final volume of the diluted (0.01 N) solution
