Answer:
403 mL
Explanation:
First, I will assume that the mole is 1, because you are not specifing this.
Now, with the innitial data, we need to get the pressure:
T = 65+273 = 338 K
V = 500 / 1000 = 0.5 L
Now if:
PV = nRT
Then:
P = nRT/V and V = nRT/P
Let's calculate the P:
P = 1 * 0.082 * 338 / 0.5 = 55.432 atm
The standard temperature is 0° C or 273 K so, the volume is:
V = 1 * 0.082 * 273 / 55.432
V = 0.40384 L or simply 403.84 mL
I think I found it. It was a cylinder
Answer:
283 grams KCl
Explanation:
To convert form moles to grams, you need to find the molar mass of your compound. The molar mass can be calculated using the individual molar masses of each element in the compound. These values can be found on the periodic table.
Molar Mass (K) = 39.098 g/mol
Molar Mass (Cl) = 35.453 g/mol
Molar Mass (KCl) = 39.098 g/mol + 35.453 g/mol
Molar Mass (KCl) = 74.551 g/mol
Now that you know the molar mass, you want to multiply it by the moles of KCl. This is possible because the molar mass is actually a ratio representing the relationship of grams and moles. In KCl, there are 74.551 grams of KCl for every 1 mole of KCl. When you are multiplying, make sure you set the ratio up in a way that allows for the units to cancel out and leave grams as your final answer.
3.80 moles KCl 74.551 g
----------------------- x ------------------- = 283 grams KCl
1 mole KCl
The final answer should have 3 sig figs because the value the problem gave (3.80 mol) also has 3 sig figs.
Answer: The final temperature of the system will be 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of steam = 25 g
= mass of water = 0.2384 kg = 238.4 g (1kg=1000g)
= final temperature = ?
= temperature of steam = 
= temperature of water = 
= specific heat of steam = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![25g\times 1.996J/g^0C\times (T_{final}-116)=-[238.4g\times 4.184J/g^0C\times (T_{final}-8)]](https://tex.z-dn.net/?f=25g%5Ctimes%201.996J%2Fg%5E0C%5Ctimes%20%28T_%7Bfinal%7D-116%29%3D-%5B238.4g%5Ctimes%204.184J%2Fg%5E0C%5Ctimes%20%28T_%7Bfinal%7D-8%29%5D)
Therefore, the final temperature of the system will be
