When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.
According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.
150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:
Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.
When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
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Answer:
108.6 g
Explanation:
- 2NaN₃(s) → 2Na(s) + 3N₂(g)
First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 0 °C ⇒ 0 + 273.2 = 273.2 K
<u>Inputting the data</u>:
- 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 2.5 mol N₂ *
= 1.67 mol NaN₃
Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:
- 1.67 mol * 65 g/mol = 108.6 g
Answer : The value of rate constant is, 
Explanation :
First we have to calculate the rate constant, we use the formula :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time passed by the sample = 20 min
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 100 - 85 = 15 g
Now put all the given values in above equation, we get
Therefore, the value of rate constant is,
The H+ concentration in a solution =10^(-1.25)=0.05623413M
