Atomic number should be the answer
Explanation:
The given data is as follows.
Volume of lake =
= 
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = 
=
mg
=
kg
Flow rate of river is 50 
Volume of water in 1 day = 
=
liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are
or 
Flow rate of sewage = 
Volume of sewage water in 1 day =
liter
Concentration of sewage = 300 mg/L
Total amount of pollutants =
or 
Therefore, total concentration of lake after 1 day = 
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence,
= 
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
The reaction of removing CO2
using LiOH is the following:
2 LiOH + CO2 -----> Li2CO3
+ H2O
By solving the amount of CO2
the LiOH can scrub:
(3.50 × 10^4 g LiOH) (1 mol LiOH/
24 g LiOH) ( 1 mol CO2 / 2 mol LiOH) ( 44 g CO2 /1 mol CO2) = 32, 083.33 g CO2
it can scrub
<span>Since number of astronaut = 32,
083.33 g / 9 (8.8 × 10^2) = 4 astronaut</span>
pH of 0.048 M HClO is 4.35.
<u>Explanation:</u>
HClO is a weak acid and it is dissociated as,
HClO ⇄ H⁺ + ClO⁻
We can write the equilibrium expression as,
Ka = ![$\frac{[H^{+}] [ClO^{-}] }{[HClO]}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BClO%5E%7B-%7D%5D%20%20%7D%7B%5BHClO%5D%7D)
Ka = 4.0 × 10⁻⁸ M
4.0 × 10⁻⁸ M = 
Now we can find x by rewriting the equation as,
x² = 4.0 × 10⁻⁸ × 0.048
= 1.92 × 10⁻⁹
Taking sqrt on both sides, we will get,
x = [H⁺] = 4.38 × 10⁻⁵
pH = -log₁₀[H⁺]
= - log₁₀[ 4.38 × 10⁻⁵]
= 4.35
The molar mass of B(NO₃)₃ - Boron nitrate : 196.822 g/mol
<h3>Further explanation</h3>
In stochiometry therein includes
<em>Relative atomic mass (Ar) and relative molecular mass / molar mass (M) </em>
So the molar mass of a compound is given by the sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
The molar mass of B(NO₃)₃ - Boron nitrate :
M B(NO₃)₃ = Ar B + 3. Ar N + 9.Ar O
M B(NO₃)₃ = 10.811 + 3. 14,0067 + 9. 15,999
M B(NO₃)₃ = 196.822 g/mol