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.
Each covalent H bond is nonpolar.
1) Write the balanced chemical equation
2HCl + Na2 CO3 ----------> 2NaCl + H2CO3
2) Write the molar ratios:
2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3
3) Convert 0.15g of sodium carbonate to number of moles
3a) Calculate the molar mass of Na2CO3
Na: 2 * 23 g/mol = 46 g/mol
C: 12 g/mol =
O: 3 * 16 g/mol = 48 g/mol
molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol
3b.- Calculate the number of moles of Na2CO3
# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3
4) Calculate the number of moles of HCl from the molar proportion:
[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl
5) Calculate the volume of HCl from the definition of Molarity
Molarity, M = # moles / volume in liters
=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters
0.028 liters * 1000 ml / liter = 28 ml.
Answer: 28 mililiters of 0.1 M HCl.
Answer:
660kcal
Explanation:
The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.
We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:
3.3L * (1000mL / L) * (5g/100mL)= 165 g.
If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.
