Answer:
compound is a material formed by chemically bonding two or more chemical elements.
Explanation:
20.21*10^-4 mg/L is the concentration.
Explanation:
Given:
Pressure=1 atm.
Temperature= 0 degrees
Volume percent of air is 21%
Henry's Law constant K = 2.28 x 10^-3 mole/L-atm
The partial pressure of oxygen is 0.21 atm.
By Henry's law:
Concentration= K X Partial pressure
= 2.28*10^-3 *0.21
= 4.79*10^-4 moles/litre
Since, at STP 1 mole of oxygen occupies volume of 22.4L
concentration =mass/volume
mass = 4.79*10^-4-4*22.4
= 20.21*10^-4 mg
20.21*10^-4 mg/L is the concentration.
Answer: The mass percentage of 
is 5.86%
Explanation:
To calculate the mass percentage of in the sample it is necessary to know the mass of the solute ( in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the 
precipitate. We can establish a relation between the mass of and using the stoichiometry of the compounds:
Since for every mole of Tl in there are two moles of Tl in , we have:
Using the molar mass of we have:
Finally, we can use the mass percentage formula:
Answer:
B.) 1.3 atm
Explanation:
To find the new pressure, you need to use Gay-Lussac's Law:
P₁ / T₁ = P₂ / T₂
In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.
P₁ = 1.2 atm P₂ = ? atm
T₁ = 20 °C + 273 = 293 K T₂ = 35 °C + 273 = 308 K
P₁ / T₁ = P₂ / T₂ <----- Gay-Lussac's Law
(1.2 atm) / (293 K) = P₂ / (308 K) <----- Insert values
0.0041 = P₂ / (308 K) <----- Simplify left side
1.3 = P₂ <----- Multiply both sides by 308
