The ions of Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
The periodic table is a systematic arrangement of elements in order of their atomic numbers into a set of 8 columns each called groups and a set of 7 rows each called a period.
Elements are arranged in different groups according to the number of Valence electrons they have.
- For instance, elements in the group I of the periodic table are highly electropositive and as such are highly reactive.
The same is evident in group 7 elements are highly electronegative and have high electron affinity and as such are unstable and reactive.
- However, Noble gases, <em>group VIII</em> elements have a full octet configuration on their outermost shell and as such are highly stable.
Consequently, the <em>Noble gases ion</em> has a stable Valence electron configuration.
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Answer:
See attached picture.
Explanation:
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In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.
In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.
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0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
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Charge is the measure of extra positive or negative particles an object has
Combined gas law is
PV/T = K (constant)
P = Pressure
V = Volume
T = Temperature in Kelvin
For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL
hence the volume of gas at stp is 2012.6 mL
