Answer: Molarity of the prepared solution is 
Explanation:
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
where,
n= moles of solute
= volume of solution in ml = 500 ml
moles of solute =
Now put all the given values in the formula of molarity, we get
Thus molarity of the prepared solution is
Answer:
The answer to your question is below
Explanation:
1)
Balanced chemical reaction
2CH₃OH + 3O₂ ⇒ 2 CO₂ + 4H₂O
Reactant Element Product
2 C 2
8 H 8
8 O 8
Molar mass of CH₃OH = 2[12 + 16 + 4]
= 2[32]
= 64 g
Molar mass of O₂ = 3[16 x 2] = 96 g
Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67
Experimental proportion CH₃OH/O₂ = 60/48 = 1.25
Conclusion
The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion
2)
Balanced chemical reaction
S₈ + 12O₂ ⇒ 8SO₃
Reactant Elements Products
8 S 8
24 O 24
Molar mass of S₈ = 32 x 8 = 256 g
Molar mass of O₂ = 12 x 32 = 384 g
Theoretical proportion S₈ / O₂ = 256 / 384
= 0.67
Experimental proportion S₈ / O₂ = 40 / 35
= 1.14
Conclusion
The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.
I think the answer might be false. I hope this helps!
Answer:
Reduction involves a half-reaction in which a chemical species decreases its oxidation number, usually by gaining electrons. The other half of the reaction involves oxidation, in which electrons are lost. Together, reduction and oxidation form redox reactions (reduction-oxidation = redox).
Explanation:
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