Answer:
![49^oC](https://tex.z-dn.net/?f=49%5EoC)
Explanation:
At
, the heat of vaporization of water is given by:
![\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bvap%7D%20%3D%2043988%20J%2Fmol%5Ccdot%20%5Cfrac%7B1%20mol%7D%7B18.016%20g%7D%20%3D%202441.6%20J%2Fg)
The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:
![Q_1 = \Delta H^o_{vap} m_w](https://tex.z-dn.net/?f=Q_1%20%3D%20%5CDelta%20H%5Eo_%7Bvap%7D%20m_w)
The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:
![Q_2 = c_{Al}m_{Al}(t_f - t_i)](https://tex.z-dn.net/?f=Q_2%20%3D%20c_%7BAl%7Dm_%7BAl%7D%28t_f%20-%20t_i%29)
According to the law of energy conservation, the heat lost is equal to the heat gained:
or:
![\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bvap%7D%20m_w%20%3D%20c_%7BAl%7Dm_%7BAl%7D%28t_f%20-%20t_i%29)
Rearrange for the final temperature:
![\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bvap%7D%20m_w%20%3D%20c_%7BAl%7Dm_%7BAl%7Dt_f%20-%20c_%7BAl%7Dm_%7BAl%7Dt_i)
We obtain:
![\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bvap%7D%20m_w%20%2B%20c_%7BAl%7Dm_%7BAl%7Dt_i%20%3D%20c_%7BAl%7Dm_%7BAl%7Dt_f)
Then:
![t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC](https://tex.z-dn.net/?f=t_f%20%3D%20%5Cfrac%7B%5CDelta%20H%5Eo_%7Bvap%7D%20m_w%20%2B%20c_%7BAl%7Dm_%7BAl%7Dt_i%7D%7Bc_%7BAl%7Dm_%7BAl%7D%7D%20%3D%20%5Cfrac%7B2441.6%20J%2Fg%5Ccdot%200.48%20g%20%2B%200.903%20%5Cfrac%7BJ%7D%7Bg%5EoC%7D%5Ccdot%2055%20g%5Ccdot%2025%5EoC%7D%7B0.903%20%5Cfrac%7BJ%7D%7Bg%5EoC%7D%5Ccdot%2055%20g%7D%20%3D%2049%5EoC)
Answer:
We are given:
Volume (V) = 0.25 L
Pressure (P) = 0.93 atm
Temperature (T) = 15.4°C OR 288.4 K
<u>Solving for the number of moles of CO₂:</u>
From the ideal gas equation:
PV = nRT
replacing the variables
0.93 * 0.25 = n (0.082)(288.4)
n = 0.00983 moles
<u>Number of molecules:</u>
Number of moles= 0.00983
number of molecules in 1 mole = 6.022 * 10²³
Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³
Number of molecules = 5.91 * 10²¹
I think <span>3H2+N2==>2NH3</span>
Answer:
a.) 22.4 L Ne.
Explanation:
It is known that every 1.0 mol of any gas occupies 22.4 L.
For the options:
<em>It represents </em><em>1.0 mol of Ne.</em>
<em />
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 20 L.
The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 2.24 L.
<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>
- So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.