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4 years ago

It is difficult to break the ionic bonds in a compound because of the...

Chemistry

2 answers:

Alexandra [31]4 years ago

8 0

Answer: 1) C. Strong attraction between ions.

2) A. High melting points

Explanation: Ionic compounds are formed by combination of positively charged ions called cations formed by less electronegative metals and negatively charged ions called anions formed by more electronegative nonmetals.

The cations and anions with opposite charges are bonded through strong coloumbic forces and thus are difficult to break and have high melting points and thus exist as solids at room temperature.

$Na:11:[Ne]3s^1$

$Na^+:10:[Ne]$

$Cl:17:[Ne]3s^23p^5$

$Cl^-:18:[Ne]3s^23p^6$

$Na^++Cl^-\rightarrow NaCl$

mestny [16]4 years ago

6 0

The answer to the first one is C. Strong attraction between ions

The answer to the second one is A. High melting points

Hope this helps :)

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Explain the relationship between modern periodic table and electronic configuration

LuckyWell [14K]

Answer:

The periodic table displays the chemical elements based on their atomic numbers, electron configurations, and chemical properties.

Explanation:

(got the photo from Ms J Chem)

7 0

3 years ago

Consider the reaction: 2BrF3(g) --> Br2(g) + 3F2(g)

riadik2000 [5.3K]

Answer : The entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

Explanation :

The given balanced reaction is,

$2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)$

The expression used for entropy change of reaction $(\Delta S^o)$ is:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]$

where,

$\Delta S^o$ = entropy change of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

$\Delta S_f^0_{(Br_2)}$ = 245.463 J/mol.K

$\Delta S_f^0_{(F_2)}$ = 202.78 J/mol.K

$\Delta S_f^0_{(BrF_3)}$ = 292.53 J/mol.K

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]$

$\Delta S^o=268.74J/K$

Now we have to calculate the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition.

From the reaction we conclude that,

As, 2 moles of $BrF_3$ has entropy change = 268.74 J/K

So, 1.62 moles of $BrF_3$ has entropy change = $\frac{1.62}{2}\times 268.74=217.68J/K$

Therefore, the entropy change of reaction for 1.62 moles of $BrF_3$ reacts at standard condition is 217.68 J/K

3 0

4 years ago

Hydrogen sulfide gas gives rotten eggs their terrible odor. Hydrogen sulfide burns in air (O2) to produce sulfide dioxide gas an

gizmo_the_mogwai [7]

Answer:

2.07 mol O₂

Explanation:

First we need to write down the species present in the chemical equation, using the information given by the exercise:

H₂S + O₂ → SO₂ + H₂O

However this equation is not balanced, so now we balance it:

2H₂S + 3O₂ → 2SO₂ + 2H₂O

Now we can use the stoichiometric ratio to calculate the moles of oxygen from the moles of sulfide dioxide:

1.38 molSO₂ * $\frac{3molO_{2}}{2molSO_{2}}$ = 2.07 mol O₂

8 0

3 years ago

What other missions had been to Mars, and what did they accomplish?

lesya692 [45]

Answer:

they stayed alive?

Explanation:

easy

5 0

2 years ago

Read 2 more answers

Element X reacts with copper to form the compounds and . In which group on the Periodic Table is element X found? A) Group 1 B)

tatuchka [14]

Copper oxide
element x is oxygen location,group 16

6 0

4 years ago