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frosja888 [35]
1 year ago
14

which ionic bond is stronger: the ionic bond between calcium ions and chloride ions in crystalline calcium chloride , or the ion

ic bond between potassium ions and chloride ions in crystalline ? explain.
Chemistry
1 answer:
monitta1 year ago
4 0

CaCl2 has a stronger ionic bond than KCl because Ca+2 is smaller than K+ in size.

<h3>What is the working of ionic bonds ?</h3>

A stable link formed by the complete transfer of the valence electron is known as an ionic bond. This kind of interaction results in the formation of positive ions, also known as cations, and negative ions, also known as anions.

  • This kind of interaction is exemplified by the ionic bond in sodium chloride, a salt. The sodium valence electron is transferred to the outer electron shell of chloride. Ionic molecules form the basis of ionic compounds.

  • The three types of bonds that frequently develop are covalent bonds between nonmetals, ionic bonds between metals and nonmetals, and metallic connections between metals.

Learn more about Ionic bond here:

brainly.com/question/13526463

#SPJ4

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The three most common states of matter on Earth are solids,blank
Nostrana [21]

Answer:

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Explanation:

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7 0
2 years ago
Consider the combustion of octane (C8H18)
sesenic [268]
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ --->  16CO₂  + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
 mass of CO₂ formed is 591.8 g
3 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0
3 years ago
In each of the following blanks, only enter a numerical value.
tatiyna

Answer:

1) 1,... 2

2) 18

3) n= 3 and I=1

Explanation:

1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons

2) the maximum number of electron is given by 2n^2= 2×3^2= 18

3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1

5 0
2 years ago
What does lithium 6 and lithium 7 look like
Marysya12 [62]
Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital. 
3 0
3 years ago
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