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.
O2 gas, where there are two Oxygen atoms which are covalently bonded together
Answer:
No, i will not use a water pipe consisting of the two metals
Explanation:
Looking at the reduction potential of the both metals, it is clear that an electrochemical cell is set up with iron as the anode and copper as the cathode.
This will make the iron to quickly corrode and eventually destroy the water pipe. It is better to have a set up in which another metal that is higher than iron in the electrochemical series is combined with it.
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
Answer:
204.8g
Explanation:
The number of moles of a substance is related to its mass and molecular mass as follows:
mole (n) = mass (m) ÷ molar mass (MM)
According to this question, 3.50 moles of sodium chloride (NaCl) is added to a food.
Molar mass of NaCl = 23 + 35.5
= 58.5g/mol
Using mole = mass/molar mass
Mass = molar mass × mole
Mass = 58.5g/mol × 3.5mol
Mass = 204.75
Mass = 204.8grams.
Therefore, 204.8grams of NaCl or common salt was added to the food.
