Answer:
150 g of potassium contained 3.8 moles of potassium.
Explanation:
Given data:
Mass of potassium = 150 g
Moles of potassium = ?
Solution:
Number of moles = mass/ molar mass
Molar mass of potassium = 39 g/mol
Now we will put the values in formula:
Number of moles = mass/ molar mass
Number of moles = 150 g/ 39 g/mol
Number of moles = 3.8 mol
150 g of potassium contained 3.8 moles of potassium.
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.
Just 7 electrons this is the answer
Anything that is moving has kinetic energy, and the faster it is moving, the more kinetic energy it has. The total kinetic energy of moving particles of matter is called thermal energy. ... That's because the particles of all matter are in constant motion and have kinetic energy.
Answer:
Option A. 9.4 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 8 L
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 343 K
Final volume (V₂) =?
V₁ / T₁ = V₂ / T₂
8 / 293 = V₂ / 343
Cross multiply
293 × V₂ = 8 × 343
293 × V₂ = 2744
Divide both side by 293
V₂ = 2744 / 293
V₂ = 9.4 L
Therefore, the final volume of the gas is 9.4 L