Molarity after dilution : 0.0058 M
<h3>Further explanation
</h3>
The number of moles before and after dilution is the same
The dilution formula
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
M₁=0.1 M
V₁=6.11
V₂=105.12
Answer:
2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)
Explanation:
A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.
The only equation that meets those conditions is the last one.
A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.
C is wrong because Li⁺ and SO₄²⁻ are not elements.
D is wrong because it shows the formation of 8 mol of Li₂SO₄.
Answer:
molar mass of methane CH4
= C + 4 H
= 12.0 + 4 x 1.008
= 12.0 + 4.032
= 16.042g/mol
7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles
6.02 x 10^23 CH4 molecules
121.43 moles CH4 are present.
Explanation:
not to certain if this is right or not.. but hope it helps!
They become more stable because they achieve a full outer shell of valence electrons with the magic number of 8.
Answer:
90g of H2O
Explanation:
2H2 + O2 —> 2H2O
First, we calculate the molar masses of H2 And H20.
Molar Mass of H2 = 2g/mol
Mass conc of H2 from the balanced equation = 2 x 2 = 4g
Molar Mass of H2O = 2 + 16 = 18g/mol
Mass conc of H2O from the balanced equation = 2x18 = 36g
From the equation,
4g of H2 produced 36g of H2O
Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O
