Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
</span>
First, calculate for the amount of heat used up for increasing the temperature of ice.
H = mcpdT
H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J
Then, solve for the heat needed to convert the phase of water.
H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J
Then, solve for the heat needed to increase again the temperature of water.
H = (18 g)(4.18 J/gK)(70 k)
H = 5266.8 J
The total value is equal to 13157.8 J
Answer: 13157.8 J
Answer:
Hi there I think the awnser is B. or A. lol but im 90% its A.
Explanation:
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Answer:
Mass= 2.77g
Explanation:
Applying
P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28
PV=nRT
NB
Moles(n) = m/M
PV=m/M×RT
m= PVM/RT
Substitute and Simplify
m= (2.09×1.13×28)/(0.082×291)
m= 2.77g
