If I have 22,9 moles of a gas at a pressure of 12.4 atm and a volume of 24 liters, what is the temperature?

Chemistry

1 answer:

Grace [21]2 years ago

8 0

Answer:

160 K, using 2 SF,

Explanation:

Ideal Gas law, rearrange to solve for T (K), answer about 158.3 round to 2SF

PV = nRT

You might be interested in

2.1 g of a hydrocarbon fuel is burned in a calorimeter that contains 280 grams of water initially at 25.00◦C. After the combusti

olga2289 [7]

Answer:

$\large \boxed{\text{933 J}}$

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

$\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}$