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.
Answer:
the answer is 5.35 i hope you get it right
The balanced equation that shows the reaction between oxalic acid and permanganate ion in an acidic medium is: 2MnO4- + 5H2C2O4 + 6H+ -> 2Mn(2+) + 10CO2 + 8H2O. Thus, 1 mole of oxalic acid reacts with 0.4 mole of permanganate ion. This was obtained using stoichiometry:
1 mol H2C2O4 x (2 mol MnO4-/ 5 mol H2C2O4) = 0.4 mol MnO4-
In this redox reaction, the permanganate is reduced to manganese(II) ion.
Answer:
<em>When an atom of hydrogen loses its electron it becomes positively charged.</em>
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