A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
Below are the steps to get the answers:
<span>1.) write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
2.) You are given everything needed to calculate
q=heat transfer=2.2*10^2, H3PO4 moles= 1.5*10^-3, NaOH moles=5.0*10^-3
3.) equation is deltaHneutraliztion=q/Moles of limiting reagent
H3PO4 is limiting reagent because lowest moles, and is used up first
4.) Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj, </span>
Answer:
Its chemical formula is NaHCO3. Its formula consists of one sodium (Na) atom, one hydrogen (H) atom, one carbon (C) atom and three oxygen (O) atoms.
Explanation:
Hope this helps
Answer: pH = 4.996
Explanation:
No of moles = molarity x volume
:• no of moles of CH3COOH = 0.1M x 0.1L
n(CH3COOH) = 0.1mol
Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.
Therefore, Moles of CH3COOH becomes,
0.1 - 0.03 = 0.07 mol
Subsequently, the moles of CH3COONa increases and becomes,
0.08 + 0.03 = 0.11 mol
Using the Hendersom-Hasselbach equation,
pH = pKa + log [Moles of conjugate÷ moles of Ch3COOH]
From literature, pKa of Ch3COOH is 4.8
Thus,
pH = 4.8 + log [0.11/0.07]
pH = 4.8 + 0.1963
pH = 4.996
36.5ml is the new concentration so, its a chemical change if there's something new