Responda:
+ 0,9kJ / mol
Explicação:
Dados os calores de combustão do enxofre monoclínico e enxofre rômbico como - 297,2 kJ / mol e - 296,8 kJ / mol, respectivamente para a variação na transformação de 1 mol de enxofre rômbico em enxofre monoclínico conforme mostrado pela equação;
S (mon.) + O2 (g) -> SO2 (g)
Uma vez que são todos 1 mol cada, a mudança na entalpia será expressa como ∆H = ∆H2-∆H1
Dado ∆H2 = -296,8kJ / mol
∆H1 = -297,2kJ / mol
∆H = -296,8 - (- 297,2)
∆H = -296,8 + 297,2
∆H = 297,2-296,8
∆H = + 0,9kJ / mol
Portanto, a mudança na entalpia da equação é + 0,9kJ / mol
If there is 4 of CH2CI2, then there is 8 of CI. There is already 2 in each one and there is 4 sets so 2x4=8
Answer:
Aluminum hydroxide Al(OH)3
Explanation:
Sodium switches places with Aluminum and the new products are formed.
A balanced equation would look like this.
AlCl3 + 3NaOH → Al(OH)3 + 3NaCl
Answer:
c. 2.16 × 10^8 kJ
Explanation:
In the given question, 2 C-12 nuclei were used for the reaction and the mass of C-12 is 12.0000 amu. Therefore, for 2 C-12 nuclei, the mass is 2*12.0000 = 24.0000 amu.
In addition, a Na-23 and a H-1 were formed in the process. The combined mass of the products is 22.989767+1.007825 = 23.997592 amu
The mass of the reactant is different from the mass of the products. The difference = 24.0000 amu - 23.997592 amu = 0.002408 amu.
Theoretically, 1 amu = 1.66054*10^-27 kg
Thus, 0.002408 amu = 0.002408*1.66054*10^-27 kg = 3.99858*10^-30 kg
This mass difference is converted to energy and its value can be calculated using:
E = mc^2 = 3.99858*10^-30 *(299792458)^2 = 3.59374*10^-13 J
Furthermore, 1 mole of hydrogen nuclei contains 6.022*10^23 particles. Thus, we have:
E = 3.59374*10^-13 * 6.022*10^23 = 2.164*10^11 J = 2.164*10^8 kJ
Answer:
pH = 3.513
Explanation:
Hello there!
In this case, since this titration is carried out via the following neutralization reaction:
We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:
Now, we can see that the moles of acid, base and acidic salt are all:
And therefore the concentration of the salt at the equivalence point is:
![[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M](https://tex.z-dn.net/?f=%5BHONH_3%5E%2BCl%5E-%5D%3D%5Cfrac%7B0.0044mol%7D%7B0.022L%2B0.0293L%7D%20%3D0.0858M)
Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:
Whereas the equilibrium expression is:
![Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHONH_2%5D%7D%7B%5BHONH_3%5E%2B%5D%7D)
Whereas Ka is computed by considering Kw and Kb of hydroxylamine:
So we can write:
And neglect the x on bottom to obtain:
And since x=[H3O+] we obtain the following pH:
Regards!
