The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.
For the reaction shown in question 7, we can divide it into half equations as follows;
Oxidation half equation;
6 Al (s) -------> 6Al^3+(aq) + 18e
Reduction half equation;
3Cr2O7^2-(aq) + 42H^+(aq) + 18e -----> 6Cr^3+(aq) + 21H2O(l)
The balanced reaction equation is;
6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq) -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)
The E° of this reaction is obtained from;
E° anode = -0.04 V
E°cathode = +1.50 V
E° cell = +1.50 V - (-0.04 V) = 1.54 V
Given that;
ΔG° = -nFE°cell
n = 3, F = 96500, E°cell = 1.54 V
ΔG° = -(3 × 96500 × 1.54)
ΔG° = -443.83KJ/mol
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Answer:
The speed of the 60.0 kg skater should be 0.281 m/s
Explanation:
<u>Step 1: </u>Data given
Mass of skater 1 = 45.0 kg
speed of skater 1 = 0.375 m/s
Mass of skater 2 = 60.0 kg
<u>Step 2:</u> Calculate the speed of skater 2
To solve this problem, we will use 'Conservation of momenton'. This means the momentum before the push equals the momentum after.
momentum p = m*v
Momentum p(before) = momentum p(after)
m1*v1 = m2 * v2
⇒ with m1 = mass of skater 1 = 45.0 kg
⇒ with v1 = the velocity of skater 1 = 0.375 m/s
⇒ with m2 = the mass of skater 2 = 60.0 kg
⇒ with v2 = the velocity of skater 2 = TO BE DETERMINED
45.0 * 0.375 = 60.0 * v2
v2 = (45.0*0.375)/60
v2 = 0.281 m/s
The speed of the 60.0 kg skater should be 0.281 m/s
Answer:
metal atoms lose electrons to form positive ions (cations ) non-metal atoms gain electrons to form negative ions (anions )
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Explanation:
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Calculate the number of mole of 5O2:
1.2 x 5/4=1.5 mol
