Answer:
1.68L
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Initial Molarity (Mi) = 7M
Initial volume (Vi) 0.6L
Final Molarity (Mf) = 2.5M
Final Volume (Vf) =.?
Step 2:
Determination of the final volume of the solution.
This can be achieved as shown below:
MiVi = MfVf
Divide both side by Mf
Vf = MiVi /Mf
Vf = 7 x 0.6 / 2.5
Vf = 1.68L
Therefore, the new volume of the solution is 1.68L
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.
Equilibrium constant,Kc = ( [C]^2 ) ÷ ( [B]^2 × [A]^3 )
Answer:
pH change is -0.07
Explanation:
Using H-H equation for acetic acid:
pH = pKa + log [Acetate salt] / [Acetic acid]
Replacing:
pH = 4.74 + log[1.188M] / [1.188M]
pH = 4.74
The HCl reacts with sodium acetate producing acetic acid, thus:
HCl + CH₃COONa → CH₃COOH + NaCl
That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.
As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:
sodium acetate: 1.188mol - 0.1mol = 1.088mol
Acetic acid: 1.188mol + 0.1mol = 1.288mol
Using again H-H equation:
pH = 4.74 + log[1.088M] / [1.288M]
pH = 4.67
pH change is: 4.67 - 4.74 = -0.07
