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.
1. air components change from place to place, while water will just stay the same
2. Water doesn't have nor show individual properties, but air does show that it has it's own property
3. You can separate different gases from air physically, but with water you must pass electricity through it.
Don't take my word for it, this is just what I learned back in middle school.
Answer:
sodium bromide (NaBr) potassium hydroxide (KOH) magnesium chloride (MgCl2) silicon dioxide (SiO2) sodium oxide (Na2O)
Explanation:
Answer:
grams of solution = 551.98 g
Explanation:
Given data:
Percentage of solution = 32.9
Mass of solute = 181.6 g
Grams of solvent = ?
Solution:
Formula:
% = [grams of solute / grams of solution] × 100
Now we will put the values in formula.
32.9 = [ 181.6 g / grams of solution] × 100
grams of solution = 181.6 g × 100 / 32.9
grams of solution = 18160 g /32.9
grams of solution = 551.98 g
