<u>Answer:</u> The mass of chlorine needed by the plant per day is 
<u>Explanation:</u>
We are given:
Volume o water treated per day = 25,000,000 gallons
Converting this volume from gallons to liters, we use the conversion factor:
1 gallon = 3.785 L
So, 
Amount of chlorine applied for disinfection = 10 mg/L
Applying unitary method:
For 1 L of water, the amount of chlorine applied is 10 mg
So, for
of water, the amount of chlorine applied will be 
Hence, the mass of chlorine needed by the plant per day is 
The oxidation number of chlorine in the Cl-<span> ion is -1.</span>
In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =

= constant

=

==> P1 * V1 = P2 * V2
V2 =

=
V2 = 0.00581 m3 = 5.81 L
Answer:The boiling point of the solution is 108° C.
Explanation:
Boiling point of pure water=T=
Boiling point of water after addition of 500 g of ethylene glycol=
Mass of water = 500g = 0.5 kg (1000 g = 1 kg)






The boiling point of the solution is 108° C.
Answer:
P2= 125.26 Kpa
Explanation:
V1= 489.6 ml=0.4896L
V2= 750 ml= 0.750L
V1= 180 KPa= 180000 Pa
P2= ?
T1= 10 = 10 + 273.15 = 283.15K
T2= 28.7+273.15= 301.85K
180000Pa* 0.4896L/ 283.15K * 301.85K/0.75L
P2= 12526.28553
P2= 125.26 KPa