C. the denser the plants the better.
Answer:
C
It’s called neutralization. The alkali has neutralized the acid by taking away the H+ ions and then turning them into water.
Answer:
Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol
Explanation:
forming CS2 means that it should in the product side
C(graphite) + O2 → CO2 ΔH = -393.5
2S(rhombic) + 2O2 → 2SO2 ΔH = -296.4 x 2
CO2 + 2SO2 → CS2 + 3O2 ΔH = -1073.6 x -1
the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.
the third reaction is reversed (multiplied by -1) so that CS2 will be on the product side.
after adding the reaction and cancelling out similarities, the final reaction is: C(graphite) + 2S(rhombic) → CS2
Add ΔH to find the enthalpy of formation of CS2
ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x -1) = 87.3 KJ/mol
be aware of signs
The concentration of A will be <em>0.34 mol·L⁻¹</em> after 60 min.
In a first-order reaction, the formula for the amount remaining after <em>n</em> half-lives is
![\text{[A]} = \frac{\text{[A]}_{0}}{2^{n}}\\](https://tex.z-dn.net/?f=%5Ctext%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%7B2%5E%7Bn%7D%7D%5C%5C)
If 
∴ ![\text{[A]} = \frac{\text{2.7 mol/L}}{2^{3.0}} = \frac{\text{2.7 mol/L}}{8.0} = \textbf{0.34 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B%5Ctext%7B2.7%20mol%2FL%7D%7D%7B2%5E%7B3.0%7D%7D%20%3D%20%5Cfrac%7B%5Ctext%7B2.7%20mol%2FL%7D%7D%7B8.0%7D%20%3D%20%5Ctextbf%7B0.34%20mol%2FL%7D)
For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.
From the problem, you know that 1) = 2.000 g.
Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).
1) = 5.022 g - 3.715 g = 1.307 g
Now you can complete our original task.
Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.
