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Zielflug [23.3K]
2 years ago
14

When a stable diatomic molecule spontaneously forms from its atoms, what are the signs of Delta H, Delta S, and Delta G? A. + +

+ B. + - - C. - - + D. - - -
Chemistry
1 answer:
Neko [114]2 years ago
3 0
H will definitely be positive because a bond is always more stable than no bond surely if it is a sigma bond.
For G you can't really know because you don't know how much energy is provided by the bond and if it outways the loss in disorder.


The reaction will become more spontaneous with a lower temperature because H tells you the reaction is exotherm
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<span>NaCH3COO (s) + HCl (aq) ---> HCH3COO (aq) + NaCl (s)</span>
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Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM
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Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will  B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: <u><em>The value of Kc remains the same.</em></u>

<u><em> </em></u>As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, <u><em>The reaction must: A. run in the forward direction to reestablish equilibrium</em></u>.

By decreasing the volume, and so that Kc remains constant, being:

Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2}  ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume}  } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume

 where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so,  the number of moles of Cl₂ should decrease.<u><em>The number of moles of Cl2 will  B. decrease.</em></u>

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, <em><u>the value of Qc C. is less than Kc.</u></em>

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2 years ago
An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th
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In a 1 M solution of NH3(aq), identify the relative molar amounts of these species. Arrange them from most to least.
Lera25 [3.4K]

Answer:

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
2 years ago
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