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4 years ago

A change where compounds are formed or decomposed is a(n)

1 answer:

6 0

It's a chemical change. In synthesis you add compounds which form new ones. This means that it cannot be a physical or phase change because you form something new. Not all synthesis and decomposition reactions use nuclear energy therefore it is a chemical change.

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Consider the following reaction where Kc = 1.80×10-2 at 698 K:

Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

$Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}$

In this case:

$Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}$

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

$[H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}$ =2.09*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}$ =4.14*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{0.280 moles}{1 Liter}$ = 0.280 $\frac{moles}{liter}$

So,

$Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }$

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

8 0

4 years ago

Which of the following factors could you change in a system and not change its induced EMF?

krok68 [10]

The options attached to the question above are listed below:
A. Magnetic field.
B. Type of wire.
C. Velocity of the wire.
D. Length of the wire in the field.

ANSWER
The correct option is B.
The factors that determine the induced current in a system are: the number of wires in the coil, the strength of the magnetic field and speed of armature rotation [speed of cutting]. Generally, the induced electromotive force across a conductor is equal to the rate at which magnetic flux is cut by the conductor. The type of wire used does not affect the induced EMF.

8 0

3 years ago

Read 2 more answers

What is the purpose of the uninoculated control tubes used in the oxidation fermentation test?

Black_prince [1.1K]

The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.

5 0

3 years ago

How do you do these?

Agata [3.3K]

Solve these problems like weighted averages:

The first one:

Multiply the masses (isotope numbers) by the decimal form of the percentage. Add them

0.076 (6) + 0.924 (7) = 6.924

The second one:

0.2 (10) + 0.8 (11) = 10.8

If you think about it, these answers make sense. 6.924 is much closer to 7 than to 6 (since there's a lot more lithium-7 than there is lithium-6). 10.8 is closer to 11 than to 10.

6 0

3 years ago

Part A of the lab involved adding 4 mL increments of distilled water to 5.00 mL of antimony trichloride solution. The antimony t

Mrrafil [7]

Answer:

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Explanation:

The total volume of the solution is:

4mL + 5.00mL + 12.0mL = 21mL

As the volume of the SbCl3 is 5.00mL, the dilution factor is:

21mL / 5.00mL = 4.2 times

The concentration of SbCl3 is:

0.10M SbCl3 / 4.2 times = 0.0238M SbCl3

The concentration of H+ = [HCl]:

4.5M / 4.2 times = 1.07M H+

The initial concentration of Cl- is:

3 times SbCl3 + HCl = 0.10M*3 + 4.5M =

3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-

4.8M Cl- / 4.2 times = 1.14M Cl-

3 0

3 years ago