Answer:
1s22s22p63s23p64s23d104p4
Explanation:
Answer:
The numbers, positive, negative signs and the symbol of elements are used to represent the ions.
Explanation:
Steps to write the ions:
- Write the symbol of an element. For example in case of magnesium we would write Mg.
- In second step write the number of electrons in superscript that are lost or gained by an atom. For example magnesium atom loses two electrons written as followed, Mg².
- In third step write the charge as superscript after the numbers. For example, magnesium loses two electrons that's why we put the positive sign after 2, because by losing the electrons cations are formed. Mg²⁺
- If the atom of an element lose or gain only one electron then 1 is omitted and only the negative or positive sign are written on superscript. For example in case of sodium cation and chlorine anion, Na⁺ , Cl⁻.
Answer:Lone pairs are the valence electron pair of any element which do not take part in bonding but affect the shape of molecules. Bond pairs or shared pairs are the electron pair which does both affect the geometry of molecules and take part in chemical bonding. These are form due to sharing of electrons.
Explanation:
Answer:
49.95 g of HCl
Explanation:
Let's formulate the chemical equation involved in the process:
Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O
This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2
1 mole Ca(OH)2 ---- 2 moles HCl
0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl
Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:
1 mole of HCl ---- 36.46094 g
1.37 moles ------ x = 49.95 g of HCl
Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
