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sammy [17]
3 years ago
9

Which processes are affected by freeze and thaw cycles

Chemistry
2 answers:
mr_godi [17]3 years ago
7 0

Answer:

A C D

Explanation:

galben [10]3 years ago
3 0

out of these choices:

creep

deposition

landslide

slump

weathering

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(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the
djyliett [7]

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 :)

3 0
3 years ago
Suppose you wanted to dissolve 40.0 g NaOH in enough H2O to make 6.00 dm3 of solution
dezoksy [38]

Molarity of solution = 1.6 M

<h3>Further explanation</h3>

Given

40 g NaOH

6 L solution

Required

Steps to solve the problem of molarity

Solution

No additional information about the question.

If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)

But if you want to know the molarity of a solution, then

  • 1. we calculate the moles of NaOH

\tt mol=\dfrac{mass}{MW}

MW(molecular weight) of NaOH=

Ar Na+ Ar O + Ar H

23 + 16 + 1 = 40 g/mol

so mol NaOH :

\tt mol=\dfrac{40~g}{40~g/mol}=1~mol

  • 2. Molarity(M)

\tt M=\dfrac{n}{V}\\\\M=\dfrac{1}{6}\\\\M=0.16

5 0
3 years ago
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