- Add 12,69 g iodine in 1 L water
- Add 20-30 g of potassium iodide, KI
Molartiy = moles of solute/Liters of solution200 mL X (1 Liter/10^2)= 0.200 Liters0.200 Liters x (0.05 M)= .01 moles of NaOH.01 moles of NaOH x (40 g NaOH/ 1 mole NaOH)= 0.4 g NaOH are required
There is 0.5 moles of NaOH per litre To calculate 0.5 molar NaOH first know the molecular weight of NaOH i.e 40 now multiply the number of moles of NaOH you have (0.5) found as above. so to find the number of grams of NaOH we needed to start with (0.5) * (40) = 20g So dissolve 20g of NaOH in one litre of the solution to prepare 0.5 molar solution
As the pH decreases, the solution becomes 10 times more acidic for each point. A solution of pH 4 is 10 times more acidic than a solution of pH 5. A solution of pH 3 is 10 times more acidic than a solution of pH 4. 10 x 10 = 100 A solution of pH 3 is 100 times more acidic than a solution of pH 5.
right i dont know this 4 sure but because u want a 0.1 mol/dm3 and u only need 100cm3 u will need 0.01mols of copper sulfate to dilute in 100cm3. soo now u have a solution that is 0.01mols per 100cm3 or 0.1 mols per 1000cm3 (dm3)
A one percent solution means 1 part beach and 99 parts water. So we can answer the question by solving the equation 1 part bleach/99 parts water=x gallons bleach/20 gallons water. You can cross multiply. This is the same as 99x=20 so x=20/99 or .20 Let's check .20 gallons bleach/ 20 gallons water=.01 which is 1% .2 gallons is the same as 3.2 cups or 25.6 fluid oz.
Dissolve 0.4 g of NaOH in 100 ml of water. Try it out. Actually it is not suitable to prepare NaOH solutions in standard flasks.It should be made in beakers & must be standardised..This is done to find the correct normality...
Prepare Thyself to Deal With a Miracle was created on 1973-01-22.
If you want to explore in detail about 'why MODX development is the optimal solution for business development', check out this article: franciscahughes.wordpress.com/2018/01/01/call-for-cms-productivity-modx-development-is-one-stop-solution/
Molartiy = moles of solute/Liters of solution200 mL X (1 Liter/10^2)= 0.200 Liters0.200 Liters x (0.05 M)= .01 moles of NaOH.01 moles of NaOH x (40 g NaOH/ 1 mole NaOH)= 0.4 g NaOH are required
644 Solution Method: 1% * 64400 = ? .01 * 64000 = 644
Clue: He didn't laugh at the boss' jokes because he was…Answer: Musty, quota, elicit, lading, and QUITTING.
This methodology is composed by six phases closely related: prepare, plan, design, implement, operate, optimize. More info on http://www.ciscozine.com/2009/01/29/the-ppdioo-network-lifecycle/ website
01/01 - filigree 01/02 - emerald 01/03 - bib 01/04 - coral 01/05 - rocker 01/06 - teardrop 01/07 - dressy 01/08 - wood 01/09 - heart 01/10 - fringe 01/11 - long 01/12 - circle 01/13 - snap 01/14 - dangle 01/15 - wired 01/16 - ribbons 01/17 - robot 01/18 - butterfly 01/19 - feather 01/20 - watch 01/21 - drops 01/22 - snake 01/23 - shiny 01/24 - birdy 01/25 - spikes 01/26 - owl 01/27 - edgy 01/28 - glitz 01/29 - link 01/30 - cool 01/31 - kitty
At that point, the time will be 01:01:01 on 01/01/01.
To find the final concentration of Cl- ions, first calculate the moles of Cl- ions from each solution. Then add the moles of Cl- ions from both solutions and divide by the total volume of the mixed solution (500 ml) to get the final concentration. Using the formula C1V1 = C2V2 where C represents concentration and V represents volume, you can determine the moles of Cl- ions from each solution.
The strength of a solution may be described as a percentage or volume, where 1% hydrogen peroxide releases 3.3 volumes of oxygen during decomposition.Thus, a 3% solution is equivalent to 10 volume and a 6% solution to 20 volume, etc. Answer taken from Wikipedia 19/01/2009
There is 0.5 moles of NaOH per litre To calculate 0.5 molar NaOH first know the molecular weight of NaOH i.e 40 now multiply the number of moles of NaOH you have (0.5) found as above. so to find the number of grams of NaOH we needed to start with (0.5) * (40) = 20g So dissolve 20g of NaOH in one litre of the solution to prepare 0.5 molar solution