150 M NaOH = 150 mole / liter
20.0 mL = 0.0200 L
150 * 0.0200 = 3 moles
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
Your results are likely to be slightly off. How much off depends on how big the air bubble was and how much titrant you actually used.
You MUST get the chemical equation for this question, which is: HCL + NaOH --> H20 + NaCl This equation is already balanced. This is a double displacement reaction with neutralization meaning water and a salt will be the products. M = n/V (molarity = number of moles/ volume other re-arrangements of this formula: n = MV V = n/M Molarity is always moles per litre, so Its good to change all given volumes to litres then convert back to ml after if you have to. 42.0ml can be converted to litres: 42.0ml = 0.042L You now have VOLUME and MOLARITY of NaOH, now used n = MV to get number of moles: 0.042 x 0.150 = 0.0063mol NaOH now you know number of moles, and a BALANCED chemical equation, so you can now do a stoicheometry conversion factor to get the number of moles for HCl. 0.0063mol NaOH x1mol HCL ÷ 1mol NaOH = 0.0063 mol (moles of NaOH cancel out and now you have # of moles for HCl. at this point you know HCL has Molarity 0.126 and 0.0063mol. now solve for volume: V = n/M V = 0.0063/0.126 V = 0.05L You now know Volume of HCL is 0.05L, now convert it mL because the question is asking for it. 0.05L = 0.05 x 1000 = 50ml Question complete! *note* in this chemical equation, the number of moles is the same because the chemical equation is a 1:1 ratio, meaning their are no coefficients in front of an substance. is the HCl or NaOH had coefficients, then the answer would be completly different, thats why you need the stoicheometry mole to mole conversion factor that I added in.
We need 3 moles of potassium perchlorate.
Four moles of potassium chlorate are needed.
Why all peoples are selfish
Methyl orange
acid-base titration
What you are doing here is titration. You know you have a solution of HCl, but you do not know how much HCl is in it. For this you use something that can react with HCl (NaOH) and use an indicator to tell you when the reaction is complete. The reaction is pretty simple: HCl + NaOH --> H2O + NaCl You can see here that NaOH and HCl have a 1:1 mol relationship. So, lets find out how many moles of NaOH you used up with concentration = moles/volume 0.10 M NaOH = moles NaOH/ 0.0197 L NaOH solution Remember that M is in moles/L moles NaOH = 0.00197 moles Since you have a 1:1 relationship of NaOH with HCl, the 0.00197 mol applies to HCl as well. The next question works the same way, but backwards. Try doing it yourself if you understood the first part before reading my answer. Find out how many moles of HCl you have so you can find out how much moles of NaOH you need for the neutralization. 0.050 M HCl = n HCl / 0.020 L HCl soln n HCl = 0.001 mol HCl Remeber the 1:1 relationship, which gives you that n NaOH = 0.001 mol. Now all you need is the volume. 0.1 M NaOH = 0.001 mol NaOH/ Volume soln V = 0.01 L = 10 mL
cuz' borax contain boric acid (H3BO3) which is very weak acid , thus it gives non accurate titration, so we add neutral glycerol to increase the acidity by the way we add glycerol after titration of NaOH not before
Phenolphthalein indicates red for acids and blue for bases.
Titrations can be used to work out the initial amount of moles of a substance (for instance the number of moles of iron in a tablet).
Phenolphtalein change colorless at pH < 8 to purple blue at pH > 8 to 10
Some methods are: absorption spectrophotometry, amperometry, titration with NaOH.
An indicator that changes its colour around pH value 5 should be used in this titration. Alternatively, instead of Thymol Blue, Methyl Orange or Methyl Red may be used.
First construct a a balanced equation for the reaction : 2Na + 2H2O --> 2NaOH + H2 From the equation, we know that the ratio of Sodium(Na) to the ratio of Sodium Hydroxide(NaOH) is the same. Ar of Na = 23g/mol Number of moles = mass / Ar = 46g / 23g/mol = 2mol (moles of Na used) Since ratio of Na to NaOH is the same, therefore there are also 2 moles of NaOH formed.
moles H2SO4 = 0.040 L x 0.250 mol/L = 0.01 moles H2SO4moles NaOH = 0.05 L x 0.200 mol/L = 0.01 moles NaOHReaction: 2NaOH + H2SO4 = Na2SO4 + 2H2OIt takes 2 moles NaOH for each 1 mole H2SO4Moles H2SO4 used up by 0.01 moles NaOH = 0.005 molesmoles H2SO4 remaining = 0.005 moles H2SO4Total volume = 50 cm3 + 40 cm3 = 90 cm3 = 0.090 LFinal [H2SO4] = 0.005 moles/0.090 L = 0.055 M (to 2 significant figures)