Answer:
m C2H5OH = 191.52 g
Explanation:
∴ %v/v C2H5OH = ( v C2H5OH / v sln)×100
⇒ 48 proof = (2) (%v/v C2H5OH)
⇒ 48/2 = %v/v C2H5OH
⇒ 24 = %v/v C2H5OH
⇒ 24/100 = v C2H5OH / v sln = 0.24
∴ v sln (gin) = 1.00 L
⇒ v C2H5OH = ( 0.24 )( 1.00 L )
⇒ v C2H5OH = 0.24 L = 240 mL
⇒ m C2H5OH = (240 mL)(0.798 g/mL)
⇒ m C2H5OH = 191.52 g
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
Answer:
The answer to your question is 58.5 g of NaCl
Explanation:
Data
mass of NaCl = ?
mass of HCl = 40 g
mass of NaOH = 40 g
Balanced chemical reaction
HCl + NaOH ⇒ H₂O + NaCl
-Calculate the molar mass of the reactants
HCl = 1 + 35.5 = 36.5 g
NaOH = 23 + 16 + 1 = 40 g
-Calculate the limiting reactant
theoretical proportion = HCl/NaOH = 36.5 / 40 = 0.9125
experimental proportion = HCl / NaOH = 40 / 40 = 1
As the experimental proportion was higher than the theoretical proportion, the limiting reactant is NaOH.
-Calculate the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5
-Calculate the experimental mass of NaCl
40 g of NaOH -------------------- 58.5 g of NaCl
40 g of NaOH ------------------- x
x = (40 x 58.5) / 40
x = 58.5 g of NaCl
Answer:
Explanation:
You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.
1. Moles of B₄H₁₀
2. Molecules of B₄H₁₀
3. Atoms of H