A, nonpoint souce pollution is "a main problem with water quality."
Answer:
CuSO₄(aq) + 2 KF(aq) = CuF₂ + K₂SO₄
Explanation:
<em>The question is missing but I think it must be about writing and balancing the equation.</em>
Let's consider the unbalanced equation for the reaction that occurs when aqueous copper (II) sulfate reacts with aqueous potassium fluoride to produce a precipitate of copper (II) fluoride (<em>I fixed a mistake here</em>) and aqueous potassium sulfate. This is a double displacement reaction.
CuSO₄(aq) + KF(aq) = CuF₂ + K₂SO₄
Since only K and F atoms are not balanced, we will get the balanced equation by multiplying KF by 2.
CuSO₄(aq) + 2 KF(aq) = CuF₂ + K₂SO₄
Answer:
The 2nd One.
Explanation:
The skin and the skin cells were active in Norman's experiment until he seperated them in which the cyptoplasm then became active.
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