. One mole of N2O4 contains ___________.

Chemistry

1 answer:

ivanzaharov [21]3 years ago

7 0

One molecule of N2O4 contains

two nitrogen
four oxygen

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3 years ago

A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00 ∘C ∘C is placed in the water

aliina [53]

Answer:

$\large \boxed{\text{28.5 g}}$

Explanation:

There are two heat flows in this process and, since energy (heat) can neither be destroyed nor created, the energy change for the system must equal zero.

Data:

For Fe, m₁ = ?; C₁ = 0.452 J°C⁻¹g⁻¹; Ti = 2.00 °C; T_f = 21.50 °C

For H₂O, m₂ = 120 g; C₂ = 4.18 J°C⁻¹g⁻¹; Ti = 22.00 °C; T_f = 21.50 °C

Calculations:

1. Temperature changes

ΔT₁ = T_f - Ti = 21.50 °C - 2.00 °C = 19.50 °C

ΔT₂ = T_f - Ti = 21.50 °C - 22.00 °C = -0.50 °C

2. Mass of steel rod

$\begin{array}{ccl}\text{Heat gained by steel rod + heat lost by water} & = & 0\\m_{1}C_{1} \Delta T_{1} + m_{2}C_{2} \Delta T_{2}& = & 0\\m_{1} \times 0.452 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\times 19.50 \, ^{\circ}\text{C} + \text{120 g} \times 4.18\text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times (-0.50)\, ^{\circ}\text{C}& = & 0\\8.814m_{1}\text{ g}^{-1} - 250.8 & = &0\\\end{array}\\$

$\begin{array}{ccl}8.814m_{1}\text{ g}^{-1} & = &250.8\\m_{1} & = & \dfrac{250.8}{\text{8.814 g}^{-1}}\\\\& = & \textbf{28.5 g}\\\end{array}\\\text{The mass of the steel rod is $\large \boxed{\textbf{28.5 g}}$}$