Your answer is
Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O I hope it is at least :P
Answer:
50 N
Explanation:
it evens out at 200 N but because she's pulling with 250 her side pulls with 50 meaning it would be a pulling force of 50
Answer:
NO2
Explanation:
Calculate moles of Nitrogen and oxygen (assume total mass is 100g as you only have percentages therefore there is 63.6g of N and 36.4g of O) This is the ratio that they are in but you need the ratio to be in whole numbers therefore divide each number of moles by 2.275 (the moles of oxygen) to calculate the empirical formula. Hope this helps!
I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.
The steps are already in the correct order.
1. Record the pH when you have added 0 mL of NaOH to your beaker containing 25 mL of HCl and 25 mL of deionized water.
2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.
3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.
4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.
5. Add NaOH one drop at a time until you reach a pH of 7.00, then record the volume of NaOH added from the buret ( at about 25 mL).
6. Record the pH of your basic HCl-NaOH solution when you have added 26.00 mL, 27.00 mL, 28.00 mL, 29.00 mL and 30.00 mL of NaOH.
7. Record the pH of your basic HCl-NaOH solution when you have added 35.00 mL, 40.00 mL, 45.00 mL and 50.00 mL of NaOH from your 50mL buret.
Answer:
900 J/mol
Explanation:
Data provided:
Enthalpy of the pure liquid at 75° C = 100 J/mol
Enthalpy of the pure vapor at 75° C = 1000 J/mol
Now,
the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.
Thus, mathematically,
The heat of vaporization at 75° C
= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C
on substituting the values, we get
The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol
or
The heat of vaporization at 75° C = 900 J/mol
