The reaction ratio of hydrogen to the ammonia produced is 3:2 hence if 3 moles of hydrogen produce 2 moles of ammonia thus mathematically,
3moleH2=2mole NH3
5moleH2=?
Thus cross-multiplying
(5*2)/3= 3⅓ moles.
Answer:
The electronic structures of pentacene/MoO3/Al and pentacene/Al were studied using in situ photoemission spectroscopy. The secondary electron cutoffs, highest occupied molecular orbitals (HOMOs) and core level changes were measured upon deposition of the pentacene and MoO3 to study the electronic structures at the interface.
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The empirical formula of this compound is equal to
.
<h3>
Empirical formula</h3>
To calculate the empirical formula of a compound, it is necessary to know the number of moles present.
Therefore, we will use the molar mass of iron and oxygen to find the amount of moles, so that:






Finally, as the empirical formula is composed of integers numbers of moles, just multiply the values by the smallest common factor to transform into an integer, so that:
O => 
Fe => 
So, the empirical formula of this compound is equal to 
Learn more about empirical formula in: brainly.com/question/1363167
Answer:
15 electrones: 1S²2S²2P⁶3S²3P³. Fósforo
27 electrones: 1S²2S²2P⁶3S²3P⁶4S²3d⁷ - Cobalto.
56 electrones: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² - Bario
49 electrones: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p¹ - Indio
Explanation:
Para llenar los orbitales electrónicos de los distintos átomos debemos hacer uso de la regla de llenado electrónico de Aufbau. Por ejemplo, para el átomo con 15 electrones, la configuración electrónica es:
1S²2S²2P⁶3S²3P³. 2+2+6+2+3 = 15 electrones
Si elemento es neutro, tiene 15 protones. Es decir, es el fósforo, P.
27 electrones:
1S²2S²2P⁶3S²3P⁶4S²3d⁷ - Cobalto.
56 electrones:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² - Bario
49 electrones:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p¹ - Indio