Answer:
The order in which the orbitals is filled is 1s-2s-2p-3s-3p-4s-3d-4p-5s-4d-5p-6s-4f-5d-6p-7s-5f-6d-7p
Explanation:
S orbitals each hold two electrons, p orbitals hold 6, d orbitals hold 10, and f orbitals hold 14. 1s has the least energy and 7 p has the greatest energy, so the order in which they are filled is listed above.
Assuming ideal behavior of the gas for a fixed amount when temperature is held constant, the pressure and volume are inversely proportional as given by the expression
P1V1 = P2V2
where the terms with subscripts of one represent the initial conditions for pressure and volume of the gas while for terms with subscripts of two represent the final conditions.
Rearranging the Boyle's law equation to calculate for the final volume V2:
V2 = P1V1 / P2
V2 = (99.7 kPa)(150 mL) / 99.8 kPa
V2 = 149.85 mL
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
<span>Reaction: CI2 + H2O ----> HCIO + HCI
Oxidations states:
The oxitation state of Cl2 = 0, because the oxidation state of an atom alone or a molucule with one kind of atom is always 0.
The
oxidation state of Cl in HClO is +1 because the oxidation state of H is
+ 1, the oxidation state of O is - 2, and the molecule is neutral, so
+1 + 1 - 2 = 0
The oxidation state of Cl in HCl is - 1, because the oxidation state of H is +1 and the molecule is neutral, so - 1 + 1 = 0.
Also,
you shall remember that when an atom increases its oxidation state is
is oxidized and when an atoms reduces its oxidations state it is
reduced.
With that you conclude that the right option is the last statement: </span>Cl
has an oxidation number of 0 in Cl2. It is then reduced to CI- with an
oxidation number of –1 in HCl and is oxidized to Cl+ with an oxidation
number +1 in HClO.
Answer:
34- Solid
35-Metals form metallic bonds releasing one or more valence electrons to form a common electron reservoir which contains free mobile elactrons who will conduct electricity well.
36 When the cation is formed electrons are removed and the effective positive neuclear charge increases and it makes the remaining electrons accumulate around the neucleous making the cationic radius smaller than that of the atom
