Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
Hi,
mL are smaller than liters. So milk should be liters, right? Yeah it should. If yiu put milk on a scale, it would go down bc it is so heavy. If it was not as heavy, it would be a different story here. Think of it that way.
Answer:4)Wind energy can harm migrating birds.
Explanation:
When the work is being done, it is likely that there is an energy being enforced and when the energy is being enforced, it is likely that the energy present is being transferred in order for the work to be able to be able to be exterted upon
Answer:
- The molar mass of the solute, in order to convert from moles of solute to grams of solute.
- The density of solution, to convert from volume of solution to mass of solution.
Explanation:
Hello,
In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:
- The molar mass of the solute, in order to convert from moles of solute to grams of solute.
- The density of solution, to convert from volume of solution to mass of solution.
For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:
Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.
Regards.
