Answer:
306.43 K
Explanation:
- Use Charle's law and rearrange formula (V1/T1=V2/T2)
- Hope this helped! Let me know if you would like me to show you step-by-step how to do these types of problems.
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.
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
Explanation:
Monitor the temperature of the water with the thermometer. Stop heating the water once it nears the boiling point of 100 degrees Celsius. Add copper(II) sulfate and stir until the heated solution is saturated. When the solution is saturated, copper(II) sulfate will not dissolve anymore
i believe its true bc ik for sure air is a homogenous mixture