Explanation:
Bond forming:
Given species:
Na - Sodium Cl - Chlorine
Na has a valence electron
Cl has 7 valence electron
To form a bond of this nature, there must be a transfer of electrons.
Here, the bond is between a metal and non-metal
The metal is sodium and the non-metal is chlorine;
Na → Na⁺ + e⁻ loss of 1 electron
2,8,1 2,8
isoelectronic with Neon
Cl + e → Cl⁻ gain of one electron
isoelectronic with Argon
Na⁺ + Cl⁻ → Na⁺.Cl⁻
This bond type is an ionic bond
Answer:
In the Lewis structure for IF5 you'll need to put a total of 12 valence electrons on the Iodine atom in order to draw the Lewis structure. Remember that Iodine (I) can hold more than eight valence electrons. For the IF5 Lewis structure, calculate the total number of valence electrons for the IF5 molecule
Explanation:
<em> I be knowing hope this helps</em>
Answer: 
Explanation: A double displacement reaction is one in which exchange of ions take place.
The compounds which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
Thus the exchange of ions take place and all the compounds are soluble so the chemical formulas are followed by the symbol (aq).
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
