Answer:
3. Oxidation-Reduction Reaction
Explanation:
Fe(s) + Cl2(aq) = Fe2+(aq) + Cl-(aq)
Fe(s) -2e- ----> Fe2+(aq) oxidation
Cl2(aq) + 2e- -----> 2Cl-(aq) reduction
Answer:
1) a) 2 C2H2 + 3O2 --> 2CO2 + 2H2O
b) Combustion Reaction
c) Bonds will be broken on the reactant side which will require energy.
d) It agrees with the law of conservation of mass because there are the same amount of molecules of each element on both sides of the equation, meaning mass was not created nor destroyed.
2) a) 2Na + Cl2 --> 2NaCl
Synthesis
b) 2Fe + O2 --> 2FeO
Synthesis
Hope this helps!
Explanation:
Answer:
The outside temperature is -45.8°C
Explanation:
When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).
V1 / T1 = V2 / T2
2.95L/298K = 2.25L / T2
(2.95L/298K ) . T2 = 2.25L
T2 = 2.25L . 298K / 2.95L
T2 = 227.2K
T°K - 273 = T°C
227.2K - 273 = -45.8°C
Answer:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.025M)
Explanation:
Hello.
In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

Whereas the law of mass action (equilibrium expression) is:
![Kc=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
That in terms of initial concentrations and reaction extent or change
turns out:
![Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7Bx%2Ax%7D%7B%28%5BHI%5D_0-2x%29%5E2%7D%5C%5C%5C%5C54.3%3D%5Cfrac%7Bx%5E2%7D%7B%280.391M-2x%29%5E2%7D)
And the solution via solver or quadratic equation is:

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.391M-2*0.183M=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.391M-2%2A0.183M%3D0.025M)
Regards.
Answer:
3.11%
Explanation:
We express a solution's volume by volume percent concentration, % v/v,
Take the ratio of the isopropyl alcohol (IPA) volume to the total volume of the solution, which is 1400 mL of water+ 45 mL of IPA,
and multiply by 100 to get the percentage: 45/(1400+45) *100 = 0.0311*100 = 3.11%