<span>KCl<span>O3</span><span>(s)</span>+Δ→KCl<span>(s)</span>+<span>32</span><span>O2</span><span>(g)</span></span>
Approx. <span>3L</span> of dioxygen gas will be evolved.
Explanation:
We assume that the reaction as written proceeds quantitatively.
Moles of <span>KCl<span>O3</span><span>(s)</span></span> = <span><span>10.0⋅g</span><span>122.55⋅g⋅mo<span>l<span>−1</span></span></span></span> = <span>0.0816⋅mol</span>
And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.
At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.
And thus, volume of gas produced = <span>22.4⋅L⋅mo<span>l<span>−1</span></span>×0.0816⋅mol≅3L</span>
Note that this reaction would not work well without catalysis, typically <span>Mn<span>O2</span></span>.
Answer:
- What is the AGⓇ of this reaction? 0.
- Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
- What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
- If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.
Explanation:
1. To calculate the delta G of a reaction given the K, we use the following equation:
ΔG°= -RT ln K.
Which gives us 0 when K is 1.
2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.
3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.
4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.
dinosaur footprint
Explanation:
A dinosaur footprint is an example of a trace fossil. A trace fossil is a type of fossil that shows the activities of organisms that lived in the past.
- Fossils are the preserved remains of organisms that lived several years ago.
- Fossils are usually found in sedimentary rocks and thick layers of ice in temperate and polar regions.
- Body fossils are the remains of the body parts of an organism that has been preserved. They can be skeletal parts, teeth, eggs e.t.c
- A trace fossil shows the preserved remains of the activities of an organism.
- They can be fingerprints, burrows and borings, feccal pellets e.t.c
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Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
