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3 years ago

8

Which of the following are examples of physical change? Copper oxidizing Iron rusting Carbon dioxide sublimating Water evaporati

ng

1 answer:

kari74 [83]3 years ago

6 0

Carbon dioxide sublimation
water evaporate

physical change is just change between solid/liquid/gas, other two are chemical changes

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kenny6666 [7]

Answer:

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Explanation:

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2 years ago

Which of the following has the greatest electronegativity difference between the bonded atoms? a. A strong acid made of hydrogen

NARA [144]

Answer: Option (d) is the correct answer.

Explanation:

Electronegativity value of hydrogen is 2.2.

Electronegativity value of chlorine is 3.16.

Electronegativity value of carbon is 2.55.

Electronegativity value of oxygen is 3.44.

Electronegativity value of nitrogen is 3.04.

Electronegativity value of sodium is 0.93.

Electronegativity value of iodine is 2.66.

Therefore, calculate the electronegativity difference between the bonded atoms as follows.

Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen

= 3.16 - 2.2

= 0.96

Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon

= 3.44 - 2.55

= 0.89

Electronegativity difference of $N_{2}$ = Electronegativity value of nitrogen - electronegativity value of nitrogen

= 3.04 - 3.04

= 0

Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium

= 2.66 - 0.93

= 1.73

So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.

Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.

5 0

3 years ago

Read 2 more answers

What are the examples of all the non metals in the first twenty element

agasfer [191]

Answer:

semimetals or metalloids.

Explanation:

5 0

3 years ago

9. At equilibrium a 2 L vessel contains 0.360

klio [65]

Answer:

Ke = 34570.707

Explanation:

H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

3 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