1) Carbon-13:
Proton-6 Neutron-7 Electron-6
2)Atomic mass of element X:
(55*10+56*20+57*70)/100=56.6
1 molecule of NaCl contains 1 sodium ion (Na+), that's why if we have 3.0 moles of.
NaCl, we have 3.0 moles of Na+.
N(ions) = n(mol) · NA.
N(ions) = 3.0 moles · 6.02·1023 = 18.06 ·1023 ions.
Answer:
Gallium, Phosphorus, Chlorine, Fluorine
Explanation:
Arrange the elements in order of increasing ionization energy. Use the periodic table to identify their positions on the table.
Drag each tile to the correct box.
Tiles
chlorinefluorinegalliumphosphorus
Sequence
Answers:
(a) 1s² 2s²2p³; (b) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²; (c) 1s² 2s²2p⁶ 3s²3p⁵
Step-by-step explanation:
One way to solve this problem is to add electrons to the orbitals one-by-one until you have added the required amount.
Fill the subshells in the order listed in the diagram below. Remember that an s subshell can hold two electrons, while a p subshell can hold six, and a d subshell can hold ten.
(a) <em>Seven electrons
</em>
1s² 2s²2p³
There are two electrons in the 2s subshell and three in the 2p subshell. The remaining two electrons are in the inner 1s subshell.
(b) <em>22 electrons
</em>
1s² 2s²2p⁶ 3s²3p⁶ 4s²3d²
There are two electrons in the 4s subshell and two in the 2p subshell. The remaining 18 electrons are in the inner subshells.
(c) <em>17 electrons</em>
1s² 2s²2p⁶ 3s²3p⁵
There are two electrons in the 3s subshell and five in the 2p subshell. The remaining 10 electrons are in the inner subshells.
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or 
)
Given : kf = 1.86 
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907 )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
