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Jet001 [13]
3 years ago
7

(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the

Chemistry
1 answer:
djyliett [7]3 years ago
3 0

Answer:

Explanation:

1. The amount of CaCO3 must be so small that  

P

CO

2

 is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full  

P

CO

2

 required for equilibrium.

3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)  

K

c

=

[

CH

3

OH

]

[

H

2

]

2

[

CO

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)  

K

c

=

[

CO

]

[

H

2

]

[

H

2

O

]

; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

Hope this helps :)

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Determine whether each of the molecules below is polar or nonpolar. Tetrahedral C C l 4 CClX4 Nonpolar Tetrahedral C H 3 O H CHX
liraira [26]

Answer:

CCl4 - Nonpolar

CH3OH - polar

NH3 - polar

CS2 - Nonpolar

Explanation:

One important thing that we should know is that polarity has to do with the presence of a resultant dipole moment in a molecule.

Dipole moment is a vector quantity, This means that its direction is also taken into account when discussing the dipole moment of molecules.

Hence, symmetrical molecules such as CS2 and CCl4 are non-polar even though they have polar bonds because their dipoles cancel out(zero resultant dipole moment).

On the other hand, NH3 and CH3OH are non-symmetrical molecules hence they possess an overall dipole moment and are polar molecules.

3 0
3 years ago
What is the percent by mass of Fluorine in Nitrogen triflouride NF3?
Kipish [7]

Answer:

The answer to your question is 80.3%

Explanation:

Data

Percent by mass of F

molecules NF₃

Process

1.- Calculate the molar mass of nitrogen trifluoride

molar mass = (1 x 14) + (19 x 3)

                    = 14 + 57

                    = 71 g

2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.

                       71 g of NF₃ ------------------ 100%

                       57 g of F   ------------------- x

                            x = (57 x 100)/71

                            x = 5700 / 71

                            x = 80.3%

3.- Conclusion

Fluorine is 80.3% by mass of the molecule NF₃

           

3 0
3 years ago
What is the percentage of lithium in lithium carbonate (Li2CO3)?
ale4655 [162]
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g


% LI = 13.88/73.89*100 = 18.78% perfectly correct.
6 0
3 years ago
Read 2 more answers
What is the simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens?
butalik [34]

The simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens is 1 O atom/ 2 H atoms.

Sucrose is a diasaccharide having the formula; C12H22O11.

This implies that there are twenty two atoms of hydrogen and eleven atoms of oxygen in a molecule of sucrose.

This gives a ratio of 2 hydrogen atoms to one oxygen atom. Therefore, the simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens is 1 O atom/ 2 H atoms.

Learn more: brainly.com/question/11347582

8 0
3 years ago
Hat are the main properties of solids (in contrast to liquids and gases)?
atroni [7]
Molecules huddle close together.
cannot form to any shape.
5 0
3 years ago
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