Answer:
3, 2, 1, 6
Explanation:
- Make a list of each element and the amount it contains.
- Multiply them out to get them equivalent.
- Hope that helped! Please let me know if you need further explanation.
Answer:
The answer is: 22, 4 liters
Explanation:
We use the formula PV=nRT
At STP: 1 atm of pressure and 273 K of temperature.
PV=nRT
V=nRT/P =1 mol x 0,082 l atm/L mol x 273 K /1 atm
V=22, 386 l
With dish soap baking soda and contact solution
Answer:
Volume of chlorine solution required is around 1.2 L
Explanation:
Volume of water in the pool = 2.0*10^4 gal
1 gal = 3.79 L
Therefore, volume of water in the pool in units of Liters would be:
Density of water = 1 g/ml = 1000 g/L
![Mass\ of\ water\ in\ the\ pool = Density *volume\\\\=1000g/L*7.4*10^{4} L = 7.4*10^{7} g](https://tex.z-dn.net/?f=Mass%5C%20of%5C%20water%5C%20in%5C%20the%5C%20pool%20%3D%20Density%20%2Avolume%5C%5C%5C%5C%3D1000g%2FL%2A7.4%2A10%5E%7B4%7D%20L%20%3D%207.4%2A10%5E%7B7%7D%20g)
The accepted concentration of chlorine = 1 g/ 10⁶ g water
Therefore amount of chlorine required to disinfect the pool water would be:
![=\frac{1\ g\ chlorine*7.4*10^{7}\ g water }{10^{6}\ g\ water } =74\ g](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%5C%20g%5C%20chlorine%2A7.4%2A10%5E%7B7%7D%5C%20g%20water%20%7D%7B10%5E%7B6%7D%5C%20g%5C%20water%20%7D%20%3D74%5C%20g)
The given solution is 6.0% w/w chlorine i.e.
6.0 g chlorine in 100 g solution
Therefore, amount of solution corresponding to 74 g chlorine would be:
![=\frac{74\ g\ chlorine*100\ g\ solution}{6.0\ g\ chlorine} =1233 g](https://tex.z-dn.net/?f=%3D%5Cfrac%7B74%5C%20g%5C%20chlorine%2A100%5C%20g%5C%20solution%7D%7B6.0%5C%20g%5C%20chlorine%7D%20%3D1233%20g)
Density of the solution = 1 g/ml
![Volume\ of\ chlorine\ solution\ required = \frac{Mass}{Density}\\\\= \frac{1233g}{1.0 g/ml} = 1.2*10^{3} ml = 1.2\ L](https://tex.z-dn.net/?f=Volume%5C%20of%5C%20chlorine%5C%20solution%5C%20required%20%3D%20%5Cfrac%7BMass%7D%7BDensity%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B1233g%7D%7B1.0%20g%2Fml%7D%20%3D%201.2%2A10%5E%7B3%7D%20ml%20%3D%201.2%5C%20L)