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In this case sodium zincate is formed.
HCl : makes it acidic. it decreases the pH NaOH : makes it alkaline. it increases the pH
1. Weigh 4 g NaOH. 2. Put this NaOH in a 1 L volumetric flask. 3. Add slowly 100 mL distilled water and stir. 4. Put the flask in a thermostat at 20 0C and maintain for 1 hour. 5. Add distilled water up to the mark. Stir vigorously. 6. Standardize the solution by titration with oxalic acid, potassium hydrogen phtalate, etc. 7. Transfer the solution in a bottle and apply a label (date, name of the operator, name of the solution, normality).
1/.12 = 8.3333333333 This is the dilution factor you need to achieve. 250ml / 8.333333 = 30ml. This is a very important concept for chemistry so it may be important to try this method. The logical way works for me.
I think it's called soluble. It is; I'll just add 'dissolution'.
Nothing.
it explodes
You can add the sodium hydroxide solution(NaOH). Because NaOH reacts with Br2(Bromine) and generates NaBr. NaBr is dissolved in water while bromobenzene is layered with water, then you could remove Bromine by a liquid separation. The reaction is Br2 + 2NaOH = NaBr + NaBrO + H2O.
The water as it was was impure, that is, it contained other impurities such as sulphur and other element. Boiling the water will remove the impurities so that when the water is added to the NaOH, any reaction will be between the water and NaOH only. If the water was not boiled, the impurities in the water could react with the water.
Add a base, such as NaOH, NH3, or LiOH
you have to add naoh because in water only sodium salt of EDTA will dissolve.
add 10 grams of NaoH into 1000 ml water, it will give you NaoH of 0.25N. As for making 1N solution you need to disolve 40 grams of NaoH into 1 litre water.
Depends on your starting concentration, if it is 1.0M NaOH and you are trying to get to 0.13M. Simply measure 7mL of 1.0M NaOH, add 93mL of distilled water and mix it and you then have 0.13M NaOH.
In this case sodium zincate is formed.
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