Answer:
Double and triple covalent bonds occur when four or six electrons are shared between two atoms, and they are indicated in Lewis structures by drawing two or three lines connecting one atom to another
Explanation:
Answer:
Molarity of Ba²⁺ is 0M
Explanation:
Some barium ion, Ba²⁺ is produced when barium acetate is dissolved (Moles of barium acetate = moles of Ba²⁺ in solution). Then, Ba²⁺ reacts with sulfate ion to produce BaSO₄, an insoluble salt. The concentration of Ba²⁺ is the initial concentration - the concentration of SO₄²⁻ in solution.
Initial moles of Ba²⁺:
Moles barium acetate (Molar mass: 255.43g/mol)
0.363g * (1mol / 255.43) = 1.421x10⁻³ moles of Ba²⁺
Moles SO₄²⁻ = Moles of ammonium sulfate:
100mL = 0.100L * (0.025mol / L) = 2.5x10⁻³ moles of SO₄²⁻.
As moles of SO₄²⁻ are higher than moles of Ba²⁺, Molarity of Ba²⁺ is 0M because all moles of Ba²⁺ reacts producing BaSO₄(s), an insoluble salt.
Answer:
Heat flows from the reactor to the water
Explanation:
The thermal energy mentioned in the description is another way to say heat. The energy that is produced by the nuclear reactions leaves the reactor and enters the water, warming it.
The passage does <em>not </em>say that heat flows in the form of electricity, but rather that the turbines turned by the steam produce electricity.
The passage does <em>not </em>say that the steam produces the heat, but rather that the boiling of the water (caused by the heat) produces steam.
Answer:
The answer to your question is Copper (II) chloride
Explanation:
Data
Limiting reactant = ?
volume of CuCl₂ = 415 ml
[CuCl₂] = 3.0 M
mass of Al = 25 g
Process
1.- Calculate the mass of CuCl₂ in solution
Molarity = moles / volume
moles = Molarity x volume
moles = 3 x 0.415
moles = 1.245
Molar mass of CuCl₂ = 63.5 + (35.5 x 2)
= 63.5 + 71
= 134.5 g
134.5 g ----------------------- 1 mol
x ----------------------- 1.245 moles
x = (1.245 x 134.5)/1
x = 167.5 g of CuCl₂
2.- Balanced chemical reaction
3CuCl₂ + 2 Al ⇒ 2AlCl₃ + 3Cu
3.- Calculate proportions
Theoretical proportion = (3 x 134.5)/(2 x 27) = 403.5/54 = 7.47
Experimental proportion = 167.5 g / 25 g = 6.7
4.- Conclusion
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CuCl₂.
122g/58.44g/mol = 2.088mol