Answer:
None
Explanation:
There are two S=O. bonds and two S-O bonds in sulfate ion lewis structure. Sulfur atom is the center atom and four oxygen atoms are located around sulfur atom. There are no lone pairs in the last shell of sulfur atom.
To determine the empirical formula for the compound that contains <span>0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:
Molar mass of Na = 23 g/mol
</span>Molar mass of S = 32 g/mol
Molar mass of O = 16 g/ mol
The number of moles is obtained using the molar mass for each element.
moles Na = 0.979 g Na/ 23 g/mol Na = 0.04256
moles S = 1.365 g Na/ 32 g/mol Na = 0.04265
moles O = 1.021 g O/ 16 g/mol Na = 0.06326
We then divide each with the smallest number of moles obtained.
Na: 0.04256/ 0.04256 = 1
S: 0.04265/ 0.04256 = 1.002 ≈ 1
O: 0.06326/ 0.04256 = 1.49 ≈ 1.5
We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.
Answer: Provide the nuclear power plant with a plan to properly dispose of and recycle the wastes.
Explanation:
The nuclear power plant is producing huge amounts of electricity which is beneficial to the economy as it pushes growth. Shutting it down or relocating it is therefore not the right solution.
The problem is the radioactive waste being produced so a solution that is specific to this problem should suffice. That solution would be the provision of the nuclear plant with plans to properly dispose of the waste.
Should this happen, the nuclear plant can still stay in the same area and keep contributing to economic growth without adversely affecting people's heath.
Answer:
20.3 % NaCl
Explanation:
Given data:
Mass of solute = 45.09 g
Mass of solvent = 174.9 g
Mass percent of solution = ?
Solution:
Mass of solution = 45.09 g + 174.9 g
Mass of solution = 220 g
The solute in 220 g is 45.09 g
220 g = 2.22 × 45.09
In 100 g solution amount of solute:
45.09 g/2.22 = 20.3 g
Thus m/m% = 20.3 % NaCl
Answer: The standard enthalpy of formation of this isomer of 
is -210.9 kJ
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of formation of 
.
![\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CO_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%2Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}](https://tex.z-dn.net/?f=-511.3kJ%2Fmol%3D%5B%288%5Ctimes%20-393.5%29%2B%289%5Ctimes%20-241.8%29%5D-%5B%28%5Cfrac%7B25%7D%7B2%7D%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%7D)
