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 metal probably increases reaction rate by either holding reactant molecules in the correct orientation to react or by weakening or breaking bonds in reactant molecules to make them more reactive.
This is an example of heterogeneous catalysis.
It is heterogeneous catalysis because the catalyst is a solid and the reactants are gases. In heterogeneous catalysis, the catalyst is in a different phase than the reactants
Explanation:
got it right :)
Answer:
Hindi ko po alam pero subukan ko pong sagutan yan
<span>V = 24.0 mL + (35.2 g)(mL/10.5g) = I think i'm not all that sure but I think its this.</span>
Volume<span> of matter </span>decreases<span> under </span>pressure<span> ... -under </span>pressure<span>, the </span>particles<span> in a </span>gas<span> are </span>forced closer together<span> ... </span>factors<span> affecting </span>gas pressure<span> ... -</span>if pressure<span> in a sealed container is </span>lower than<span> outside, </span>gas will<span> rush in ...</span>
