We can use this equation for boiling point elevation:
ΔT(b) = i K(b) M
when Δ T(b) is the increase of boiling point of the solution.
and i is ( vant Hoff factor, the number of particles or ions per mole-clue.
and K(b) is boiling point increase constant for the solution ( and for water it is equal 0.52 C° Kg/mol)
We can assume i (vant Hoff factor ) = 1 as the sucrose is nonelectrolyte (not readily ionize).
So for water: Tb° = 100 c° and Kb = 0.52 c° Kg / mol
By substitute at:
ΔTb = i Kb M
∴ = 1 * 0.52 * 3.60 = 1.8432 C°
and when Tb = Tb° + ΔTb
∴ Tb = 100 + 1.8432 = 101.8432 C°
Kinetic energy would be the answer I <em>believe</em><span>. </span>The faster<span> a </span>particle moves<span>, </span>the<span> more kinetic </span>energy it<span> has.</span>
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.
The balanced neutralization equation is:
NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)
- Step 1: Calculate the reacting moles of KHP.
0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.
0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol
- Step 2: Determine the reacting moles of NaOH.
The molar ratio of NaOH to KHP is 1:1.
1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH
- Step 3: Calculate the molarity of NaOH.
1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.
[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Learn more about titration here: brainly.com/question/4225093
Answer:
25.907°C
Explanation:
In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C
The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.
Thus:

= heat of combustion of benzoic acid
= heat energy released to water
= heat energy released to the calorimeter
Therefore,
![-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})](https://tex.z-dn.net/?f=-m_%7Bcombust%7D%2AH_%7Bcombust%7D%20%3D%20%5Bm_%7Bwater%7D%2Ac_%7Bwater%7D%20%2B%20C_%7Bcalori%7D%5D%2A%28T_%7Bf%7D%20-%20T_%7Bi%7D%29)
1.056*26.42 = [0.987*4.18 + 6.66](
- 23.32)
27.8995 = [4.12566+6.660](
- 23.32)
(
- 23.32) = 27.8995/10.7857 = 2.587
= 23.32 + 2.587 = 25.907°C
Copper oxide(solid) + Sulphuric Acid (aqueous)-> Copper Sulphate (aqueous)+ Water(liquid)
In equation form:
CuO +H2SO4 -> CuSO4 + H2O
The colour change you will see is black to blue as Copper oxide is usually found as a black powder. Upon the reaction with sulphuric acid it will change to a cyan blue.
If you heat the made solution of copper sulphate, the water will evaporate and you will be left with white anhydrous copper sulphate crystals.