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3 years ago

Sandra claims that invisible fairies live above the clouds where no one can see, hear or feel them. Which of the

Chemistry

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer: D

Explanation: It's not scientific. branlist plz

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3 years ago

What is the normality of a solution containing 14.8 g of Ca(OH)₂ in 250.0 mL?

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Answer:

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3 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

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:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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3 years ago

What is the element presented, in this compound?

DerKrebs [107]

compound name compound formula elements present

water H2O hydrogen (H) and oxygen (O)

ammonia NH3 nitrogen (N) and hydrogen (H)

carbon monoxide CO carbon (C) and oxygen (O)

carbon dioxide CO2 carbon (C) and oxygen (O)

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3 years ago

Read 2 more answers

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

Oliga [24]

Answer:

K = [ HOCl ] . [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2

Explanation:

The law of Mass Action states that, at constant temperature, the rate of reaction is proportional to the active masses of each of the reactants.

The reaction above is a reversible reaction and the law of mass action also applies to it.

The rate of reaction from left-to-right reaction = r1 = k. [Cl2]^2 [H2O] [HgO]^2

Rate of reaction from right - to - left r2 = k. [hocl]^2 [HgO . hgcl2]

Then at equilibrium,

r1 = r2

k1/k2 = [HOCl ]^2 [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2 = K

where K is the equilibrium constant for the reaction.

5 0

3 years ago