Sucrose can be hydrolyzed to glucose and fructose by inventase enzyme (Fig.6.11).
Answer:
2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)
0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x
i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L
Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)
10.5 = x*x/(0.205-2x)^2
=> 10.5(4x^2-0.82x+0.042) = x^2
=>42x^2-8.61x+0.441=x^2
=>41x^2-8.61x+0.441 = 0
This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121
The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.
Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L
Answer:
The answer is Potential energy
Answer:
Explanation:
There are three heat transfers involved.
heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0
q₁ + q₂ + q₃ = 0
m₁ΔH + m₂C₂ΔT + C_calΔT = 0
Data:
m₁ = 2.1 g
m₂ = 280 g
Ti = 25.00 °C
T_f = 26.55 °C
Ccal = 92.3 J·°C⁻¹
Calculations:
Let's calculate the heats separately.
1. q₁
q₁ = 2.1 g × ΔH = 2.1ΔH g
2. q₂
ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C
q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J
3. q₃
q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J
4. ΔH
