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maria [59]
3 years ago
15

1. Fill in the blanks. Example: A nitrogen atom takes on a 3- charge when it forms an anion and becomes nitride. a. A magnesium

atom takes on a ___ charge when it forms a(an) ____ion and becomes . b. A chlorine atom takes on a ___ charge when it forms a(an) ____ion and becomes . c. An oxygen atom takes on a ___ charge when it forms a(an) ____ion and becomes . d. A potassium atom takes on a ___ charge when it forms a(an) ____ion and becomes .
Chemistry
1 answer:
Tema [17]3 years ago
3 0

Answer: a. +2, cation and magnesium ion .

b. -1,  anion, chloride

c. -2, anion, oxide

d. +1. cation , potassium ion

Explanation:

When an atom accepts an electron negative charge is created on atom and is called as anion.

When atom loses an electron positive charge is created on atom and is called as cation.

Magnesium (Mg) with atomic number of 12 has electronic configuration of 2,8,2 and thus it can lose 2 electrons to form Mg^{2+} cation and becomes magnesium ion.

Chlorine (Cl) with atomic number of 17 has electronic configuration of 2,8,7 and thus it can gain 1 electron to form Cl^{-} anion and becomes chloride.

Oxygen (O) with atomic number of 8 has electronic configuration of 2,6 and thus it can gain 2 electrons to form O^{2-} anion and becomes oxide.

Potassium (K) with atomic number of 19 has electronic configuration of 2,8,8,1 and thus it can lose 1 electron to form K^{+} cation and becomes potassium ion.

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What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
Brut [27]
Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

 H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

 HCO3-(aq) + H2O(l) → CO3-2 (aq)  + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s)  +  H+(aq)  → Ca2+(aq)   + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l)  Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq)  +  CO3-2(aq)   Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq)  ↔ Ca2+ (aq) + HCO3-(aq)   K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
6 0
4 years ago
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