Answer:
1s2 2s2 2p6 3s2 3p6 4s2 3d5
Explanation:
According to the Aufbau principle, electrons are filled in orbitals in order of increasing energy. The energy of orbitals in the electronic configuration of manganese increases from left to right, hence 3d orbital is much greater in energy than a 3p orbital.
The arrangement of orbitals in order of increasing energy is shown in the answer above.
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<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
Answer:
P' = 41.4 mmHg → Vapor pressure of solution
Explanation:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')
Xm = Mole fraction for solute (Moles of solvent /Total moles)
Firstly we determine the mole fraction of solute.
Moles of solute → Mass . 1 mol / molar mass
20.2 g . 1 mol / 342 g = 0.0590 mol
Moles of solvent → Mass . 1mol / molar mass
60.5 g . 1 mol/ 18 g = 3.36 mol
Total moles = 3.36 mol + 0.0590 mol = 3.419 moles
Xm = 0.0590 mol / 3.419 moles → 0.0172
Let's replace the data in the formula
42.2 mmHg - P' = 42.2 mmHg . 0.0172
P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)
P' = 41.4 mmHg
Answer:
After 5 second 25% C-15 will remain.
Explanation:
Given data:
Half life of C-15 = 2.5 sec
Original amount = 100%
Sample remain after 5 sec = ?
Solution:
Number of half lives = T elapsed / half life
Number of half lives = 5 sec / 2.5 sec
Number of half lives = 2
At time zero = 100%
At first half life = 100%/2 = 50%
At second half life = 50%/2 = 25%
Thus after 5 second 25% C-15 will remain.
