Answer:
![\large \boxed{\text{1763 psi}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B1763%20psi%7D%7D)
Explanation:
We can use Dalton's Law of Partial Pressures:
Each gas in a mixture of gases equals its pressure independently of the other gases
![\begin{array}{rcl}p_{\text{NO2}} + p_{\text{CO2}} & = & p_{\text{tot}} \\p_{\text{NO2}} + \text{795 psi} & = &\text{2558 psi} \\p_{\text{NO2}} & = &\text{2558 psi - 795 psi} \\& = & \textbf{1763 psi}\\\end{array}\\\text{The partial pressure of nitrogen dioxide is $\large \boxed{\textbf{1763 psi}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7Dp_%7B%5Ctext%7BNO2%7D%7D%20%2B%20p_%7B%5Ctext%7BCO2%7D%7D%20%26%20%3D%20%26%20p_%7B%5Ctext%7Btot%7D%7D%20%5C%5Cp_%7B%5Ctext%7BNO2%7D%7D%20%2B%20%5Ctext%7B795%20psi%7D%20%26%20%3D%20%26%5Ctext%7B2558%20psi%7D%20%20%5C%5Cp_%7B%5Ctext%7BNO2%7D%7D%20%26%20%3D%20%26%5Ctext%7B2558%20psi%20-%20795%20psi%7D%20%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B1763%20psi%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20partial%20pressure%20of%20nitrogen%20dioxide%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B1763%20psi%7D%7D%24%7D)
Eg: 2H2 + O2 ➡ 2H2O
This shows that sum of reproducts = sum of products.
Therefore, mass is conserved.
I hope this helps you.
Answer:
1.2 × 10⁴ cal
Explanation:
Given data
- Initial temperature: 80 °C
We can calculate the heat released by the water (
) when it cools using the following expression.
![Q_w = c \times m \times (T_f - T_i)](https://tex.z-dn.net/?f=Q_w%20%3D%20c%20%5Ctimes%20m%20%5Ctimes%20%28T_f%20-%20T_i%29)
where
c is the specific heat capacity of water (1 cal/g.°C)
![Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal](https://tex.z-dn.net/?f=Q_w%20%3D%20%5Cfrac%7B1cal%7D%7Bg.%5C%C2%B0C%7D%20%20%5Ctimes%20300g%20%5Ctimes%20%2840%5C%C2%B0C%20-%2080%5C%C2%B0C%29%20%3D%20-1.2%20%5Ctimes%2010%5E%7B4%7D%20cal)
According to the law of conservation of energy, the sum of the heat released by the water (
) and the heat absorbed by the reaction (
) is zero.
![Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal](https://tex.z-dn.net/?f=Q_w%20%2B%20Q_r%20%3D%200%5C%5CQ_r%20%3D%20-Q_w%20%3D%201.2%20%5Ctimes%2010%5E%7B4%7D%20cal)
Answer:
![V_2=313.71\ cm^3](https://tex.z-dn.net/?f=V_2%3D313.71%5C%20cm%5E3)
Explanation:
Given that,
Initial volume, ![V_1=525\ cm^3](https://tex.z-dn.net/?f=V_1%3D525%5C%20cm%5E3)
The pressure changes from 73.2 kPa to 122.5 k.Pa.
We need to find the new volume occupied by the air. Let it is V₂. It can be calculated using Boyle's law such that,
![P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{73.2\times 525}{122.5}\\\\V_2=313.71\ cm^3](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2%5C%5C%5C%5CV_2%3D%5Cdfrac%7BP_1V_1%7D%7BP_2%7D%5C%5C%5C%5CV_2%3D%5Cdfrac%7B73.2%5Ctimes%20525%7D%7B122.5%7D%5C%5C%5C%5CV_2%3D313.71%5C%20cm%5E3)
So, the new volume is
.