Question

a small bubble rises from the bottom of a lake, where the temperature and pressure are 8°c and 6.4 atm, to the waters surface, where the temperature is 25°c and the pressure is 1.0 atm. calculate the final volume (in ml) of the bubble if its initial volume was 2.1 ml. integer, decimal, or e notation allowed question 12 chemical reactions in a bomb calorimeter occur at constant pressure. a true b false clear selection question 13 a quantity of 1.435 g of naphthalene, was burned in a constant - volume bomb calorimeter. consequently, the temperature of the water rose from 20.28°c to 25.95°c if the heat capacity of the bomb plus water was, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion. a - 5.151×10³ kj/mol b - 5.151×10⁻³ kj/mol c 5.151×10⁻³ kj/mol

Explanation:

Step1: Convert temperatures to Kelvin

The formula for converting Celsius to Kelvin is \(T(K)=T(^{\circ}C)+273.15\).
For the initial temperature \(T_1 = 8^{\circ}C\), \(T_1=8 + 273.15=281.15K\).
For the final temperature \(T_2 = 25^{\circ}C\), \(T_2=25+ 273.15 = 298.15K\).

Step2: Use the combined gas law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\).
We need to solve for \(V_2\), so \(V_2=\frac{P_1V_1T_2}{P_2T_1}\).
Given \(P_1 = 6.4atm\), \(V_1=2.1mL\), \(T_1 = 281.15K\), \(P_2 = 1.0atm\), \(T_2=298.15K\).
Substitute the values: \(V_2=\frac{6.4\times2.1\times298.15}{1.0\times281.15}\).
First, calculate the numerator: \(6.4\times2.1\times298.15=(6.4\times2.1)\times298.15 = 13.44\times298.15\approx4017.14\).
Then divide by the denominator: \(V_2=\frac{4017.14}{281.15}\approx14.3mL\).

Answer:

\(14.3\)