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telo118 [61]
3 years ago
8

Sam prepares a solution of sodium chloride with a concentration of 8 g/dm³. If he only used 2g of sodium chloride, what volume o

f water did he dissolve it in, in cm³?
Chemistry
1 answer:
adoni [48]3 years ago
4 0

The volume of water he dissolved the solute is 250 cubic centimeter.

<u>Explanation</u>:

            Concentration = mass of solute in g / volume in dm^3

The concentration of a solution is defined as the ratio of mass of solute in grams to the volume of water in dm^3.

If concentration of a solution = 8 g/dm^3.

mass of solute =2 g.

Volume of a water = 2 / 8 = 0.25 dm^3. (or) 250 cubic centimeter.

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3 years ago
Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el
marysya [2.9K]

Answer:

80.92, Krypton

Explanation:

<u>What is effusion?</u>

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

<u>Graham's law of effusion of </u><u>gas</u>

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

\boxed{ \frac{Rate_1}{Rate_2} =  \sqrt{ \frac{M_2}{M_1} } }

<u>Calculations</u>

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

\frac{Rate_1}{Rate_2} = 1.7

1. 7 =  \sqrt{ \frac{M_2}{28} }

Square both sides:

2.89  = \frac{M_2}{28}

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

<u>Identity of </u><u>gas</u>

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

7 0
2 years ago
(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
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