The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay
mass = 0.37 mg = 0.37 * 10⁻³ g
molar mass = 206 g/mol
number of moles = 0.37 * 10⁻³ g/206 g/mol
number of moles of Pb-206 = 1.79 * 10⁻⁶ moles
mass = 0.95 mg = 0.95 * 10⁻³ g
molar mass = 238 g/mol
number of moles = 0.95 * 10⁻³ g/238 g/mol
number of moles = 3.99 * 10⁻⁶ moles
Assuming that all the Pb-206 were formed from U-238
Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles
Initial moles of U-238 = 5.78 * 10⁻⁶ moles
One mole of U-238 contains = 6.02 * 10²³ atoms
5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms
Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms
Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms
Learn more about Half-life at: brainly.com/question/4702752
Here's the answer, I remember doing this problem last year.
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When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.
1) Calculate q for the reaction. You must show your work.
2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.
3) Calculate ΔH for the reaction in kJ/mol. You must show your work.
Answer:
Explanation:
NH₄NO₃ = NH₄⁺ +NO₃⁻
heat released by water = msΔ T
m is mass , s is specific heat and ΔT is fall in temperature
= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )
= 1149.5 J .
This heat will be absorbed by the reaction above .
q for the reaction = + 1149.5 J
2 )
molecular weight of NH₄NO₃ = 80
No of moles reacted = 5/80 = 1 / 16 moles.
3 )
5 g absorbs 1149.5 J
80 g absorbs 1149.5 x 16 J
= 18392 J
= 18.392 kJ.
= + 18.392 kJ
ΔH = 18.392 kJ / mol
1s2 2s2 2p6 3s2 3p2
1s2 2s2 2p6 3s2 3p6 4s2 3d4
1s2 2s2 2p6
1s2 2s2 2p6 3s2 3p6 4s2 3d6
1s2 2s2 2p6 3s2
Answer:
The number of electrons for the Mg atom are 12 electrons. The electron configuration of magnesium is,
Mg (Z= 12) = 1s2 2s2 2p6 3s2
Explanation:
The first two electrons is placed in the 1s orbital. The 1s orbital can accommodate two electrons.
The next 2 electrons for magnesium go in the 2s orbital.
The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.
We’ll put six in the 2p orbital and then put the remaining two electrons in the 3s.
Therefore, the Magnesium electron configuration will be 1s22s22p63s2.
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Explanation:
Starting moles of ethanol acid = 0.020 mol
At the equilibrium 50 % of the ethanol acid molecules reacted
∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %
= 0.010 mol
Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol
Moles of the product gas formed are calculated as
0.010 mol CH3COOH * 1 mol / 2 mol CH3COOH
= 0.005 mol
Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol
That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Now Calculate the pressure :
0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas
P1/n1 = P2/n2
P2 = P1*n2 / n1
= 0.74 atm * 0.015 mol / 0.020 mol
= 0.555 atm
Answer:
Nice pic there
Explanation:
No need
Answer:
To instill a high degree of personal honor, self-reliance, and confidence in each cadet by presenting a military environment in which cadets will be forced to rely upon themselves and their shipmates to study, work, and learn.