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2 years ago

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Which statement best explains how the models differ? Model A shows the formation of ions, and model B does not show the formatio

n of ions. Model A does not show the formation of ions, and model B shows the sharing of electrons. Model A shows the sharing of electrons, and model B shows the formation of ions. Model A shows the transfer of an electron, and model B shows the sharing of electrons.

2 answers:

creativ13 [48]2 years ago

8 0

Answer) Model A shows the sharing of electrons, and model B shows the formation of ions.

enot [183]2 years ago

8 0

Answer: C

Explanation:

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Identify the single displacement reaction. C4H12 + 7O2 ⟶ 6H2O + 4CO2 2H2 + O2 ⟶ 2H2O Al2S3 ⟶ 2Al + 3S Cl2 + 2KBr ⟶ 2KCl + Br2

Varvara68 [4.7K]

Answer:- $Cl_2+2KBr\rightarrow 2KCl+Br_2$

Explanations:- The first reaction is combustion reaction as the hydrocarbon is burned in presence of oxygen to give carbon dioxide and water.

Second reaction is the synthesis reaction as in general a synthesis reaction looks like:

$A+B\rightarrow AB$

Third reaction is the decomposition reaction as it looks opposite of synthesis reaction.

$AB\rightarrow A+B$

In general, a single displacement reaction looks like:

$AB+C\rightarrow AC+B$

Fourth reaction is single replacement reaction as bromine is replaced by chlorine.

So, the correct choice is the last reaction, $Cl_2+2KBr\rightarrow 2KCl+Br_2$ .

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3 years ago

How many grams of Sg is required to produce 83.10 g SF6? S: +24F-->8SF

ozzi

Answer : The mass of $S_8$ required is 18.238 grams.

Explanation : Given,

Mass of $SF_6$ = 83.10 g

Molar mass of $SF_6$ = 146 g/mole

Molar mass of $S_8$ = 256.52 g/mole

The balanced chemical reaction is,

$S_8+24F_2\rightarrow 8SF_6$

First we have to determine the moles of $SF_6$ .

$\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles$

Now we have to determine the moles of $S_8$ .

From the balanced chemical reaction we conclude that,

As, 8 moles of $SF_6$ produced from 1 mole of $S_8$

So, 0.569 moles of $SF_6$ produced from $\frac{0.569}{8}=0.0711$ mole of $S_8$

Now we have to determine the mass of $S_8$ .

$\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8$

$\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g$

Therefore, the mass of $S_8$ required is 18.238 grams.

7 0

3 years ago

Place the following in order of increasing acid strength. HBrO2 HBrO3 HBrO HBrO4 A) HBrO2

jarptica [38.1K]

Answer:

HBrO4 < HBrO3 < HBrO2 < HBrO

3 0

3 years ago

If the density of

Soloha48 [4]

It’s the doubles I think not sure let me know if it’s true

8 0

3 years ago

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr

VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

5 0

3 years ago