When a 5.00-g sample of rbbr is dissolved in water in a calorimeter?
1 answer:
To answer this question, I found a similar problem as shown in the picture below. Let's use the given information: Heat capacity = 3.565 kJ/K, ΔT = 0.26 K. Also, the molar mass of RbBr is 165.372 g/mol.
The solution is as follows:
Molar Heat of Solution = (3.565 kJ/K)(0.26 K)(1/ 5 g)(165.372 g/mol)
<em>Molar Heat of Solution = 30.66 kJ/mol</em>
The solution is as follows:
Molar Heat of Solution = (3.565 kJ/K)(0.26 K)(1/ 5 g)(165.372 g/mol)
<em>Molar Heat of Solution = 30.66 kJ/mol</em>
You might be interested in
Answer: -
100 mm Hg
Explanation: -
P 1 =400 mm Hg
T 1 = 63.5 C + 273 = 336.5 K
T 2 = 34.9 C + 273 = 307.9 K
ΔHvap = 39.3 KJ/mol = 39.3 x 10³ J mol⁻¹
R = 8.314 J ⁻¹K mol⁻¹
Now using the Clausius Clapeyron equation
ln (P1 / P2) = ΔHvap / R x (1 / T2 - 1 / T1)
Plugging in the values
ln (400 mm/ P₂) = (39.3 x 10³ J mol⁻¹ / 8.314 J ⁻¹K mol⁻¹) x ( -
= 1.38
P₂ = 100 mm Hg