Which of the following is a strong acid?
1 answer:
You might be interested in
Given that, an experiment to measure the enthalpy change for the reaction of aqueous copper(II) sulfate, CuSO4(aq) and zinc, Zn(s) was carried out in a coffee cup calorimeter; the heat of the reaction in the whole system is calculated to be 2218.34 kJ
Heat of reaction (i.e enthalpy of reaction) is the quantity of heat that is required to be added or removed when a chemical reaction is taken place in order to maintain all of the compounds present at the same temperature.
The formula used to calculate the heat of the reaction can be expressed as follows:
Q = mcΔT
where:
- Q = quantity of heat transfer
- m = mass
- c = specific heat of water = 4.18 kJ/g °C (constant)
- ΔT = change in temparature
From the information given:
- The initial temperature (T₁) = 25° C
- The final temperature (T₂) = 91.5° C
∴
The change in temperature i.e. ΔT = T₂ - T₁
ΔT = 91.5° C - 25° C
ΔT = 66.5° C
The number of moles of CuSO₄ = 1.00 mol/dm³ × 50.0 cm³
= 0.05 moles
- Since the molar mass of CuSO₄ = 159.609 g/mol
Then;
Using the relation:
By crossing multiplying;
mass of CuSO₄ = number of moles of CuSO₄ × molar mass of CuSO₄
mass of CuSO₄ = 0.05 moles × 159.609 g/moles
mass of CuSO₄ = 7.9805 grams
∴
Using the formula from above:
Q = mcΔT
Q = 7.9805 g × 4.18 kJ/g °C × 66.5° C
Q = 2218.34 kJ
Therefore, we can conclude that the heat of the reaction is 2218.34 kJ
Learn more about the chemical reaction here:
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K =
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M