They would not survive in a natural setting!
Answer:
0.435 M
Explanation:
In case of dilution , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = initial concentration ,
V₁ = initial volume ,
M₂ = final concentration , i.e. , concentration after dilution ,
V₂ = final volume .
from , the question ,
M₁ = 0.725 M
V₁ = 300 mL
M₂ = ?
V₂ = 300 mL + 200 mL = 500 mL
Since, the final volume of solution would be the summation of the initial and final volume.
Using the above formula , the molarity of the final solution after dilution , can be calculated as ,
M₁V₁ = M₂V₂
0.725 M * 300 mL = M₂ * 500mL
M₂ = 0.435 M
1 magnesium atom, 1 sulfur atom, 4 oxygen atoms.
I believe the end result is still 83 moles since there is never an amount of sulfur atoms added to the initial amount, but rather oxygen and water is repeatedly added to it. To find it's weight, first find the molar mass of H2SO4:
H2 + S + O4 = 2.00 + 32.1 + 64.0 = 98.1 g/mol
and mass = (98.1 g/mol)(83 mol) = 8142.3 g
rounded to 8.1 x 10^3 g assuming 100% yield?
Answer:
Calcium oxide: CaO
Copper (II) nitrate: Cu(NO₃)₂
Ammonium chloride: NH₄Cl
Magnesium sulfate: MgSO₄
Copper (I) oxide: Cu₂O
Calcium chloride: CaCl₂
Explanation:
Hello there!
In this case, according the IUPAC rules for naming this inorganic salts, whereas calcium, copper (II), ammonium, magnesium and copper (I) are the cations and oxide, nitrate, chloride and sulfate are the anions, we proceed as follows:
Calcium oxide: CaO
Copper (II) nitrate: Cu(NO₃)₂
Ammonium chloride: NH₄Cl
Magnesium sulfate: MgSO₄
Copper (I) oxide: Cu₂O
Calcium chloride: CaCl₂
Best regards!
