Answer:
498 kj/mol
Explanation:
Chemical reactions occur as a result of bond breaking and bond formation.
The bonds in reactants are broken and atoms are rearranged to form new bonds.
During bond breaking energy is absorbed to break the bonds of reactants while bond formation involves the release of energy during the formation of new bonds.
In our case;
In 1 mole of the Oxygen molecule, there is one O=O bond
Energy absorbed to break O=O is 498 kJ/mol
Therefore, the ΔH required to break all the bonds in one mole of Oxygen(O₂) molecules is 498kJ/mol.
Note that, bond breaking is endothermic since energy is absorbed from the surroundings.
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Answer:
9.15×10²³molecules
Explanation:
moles=number of particles/Avogadro's number
1.52=x/6.02×10²³
by cross multiplication;
x=1.52×6.02×10²³
=9.15×10²³
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Answer:
-2.3 ºC
Explanation:
Kf (benzene) = 5.12 ° C kg mol – 1
1st - We calculate the moles of condensed gas using the ideal gas equation:
n = PV / (RT)
P = 748/760 = 0.984 atm
T = 270 + 273.15 = 543.15 K
V = 4 L
R = 0.082 atm.L / mol.K
n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol
Then, you calculate the molality of the solution:
m = n / kg solvent
m = 0.088 mol / 0.058 kg = 1.52mol / kg
Then you calculate the decrease in freezing point (DT)
DT = m * Kf
DT = 1.52 * 5.12 = 7.8 ° C
Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:
T = 5.5 - 7.8 = -2.3 ºC
The correct answer would be air
Answer:
6.25%
Explanation:
Given data:
Half life of lutetium-117 = 6.75 days
Percentage remaining after 27 days = ?
Solution;
Number of half lives = Time elapsed / half life
Number of half lives = 27 days / 6.75 days
Number of half lives = 4
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
At 3rd half life = 25%/2 = 12.5%
At 4th half life = 12.5%/2 = 6.25%