Answer:
81 °C
Explanation:
I don’t know, I just know :)
A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.
<h3>What is "ppm"?</h3>
"ppm" of "parts per million" is a unit of concentration equivalent to milligrams of solute per liters of solution.
A pool must maintain a chlorine concentration of 3.0 ppm (3.0 mg/L). The mass of chlorine in 3.4 × 10⁶ L is:
3.0 mg Cl₂/L × 3.4 × 10⁶ L = 1.0 × 10⁷ mg Cl₂
A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.
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Answer : The chemical formula for the compound is, 
Explanation :
When the element 'M' react with the 
to give .
The balanced chemical reaction is,
In this reaction, 'M' is in mono-atomic form and is in diatomic form.
By the stoichiometry,
2 moles 'M' react with the 1 mole of to give 2 moles of .
Therefore, the chemical formula of the compound is,
I would say that it's C. Seasonal temperatures have dipped over time, but I could easily be wrong, since it's my opinion. A weather condition is defined as the atmospheric conditions that 'comprise the state of the atmosphere in terms of temperature and wind and clouds and precipitation. But I believe it could just as easily be B. 150mm of rainfall is a normal average in the city.
<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
