Answer:
![m_{FeCl_2}=0.652gFeCl_2](https://tex.z-dn.net/?f=m_%7BFeCl_2%7D%3D0.652gFeCl_2)
Explanation:
Hello there!
In this case, according to the given information and the chemical reaction, whereby iron and hydrochloric acid react in a 1:2 mole ratio, it is firstly necessary to calculate the moles of iron (II) chloride from each reactant in order to figure out the limiting reactant:
![n_{FeCl_2}=2.45gFe*\frac{1molFe}{55.845gFe}*\frac{1molFeCl_2}{1molFe}=0.0439molFeCl_2\\\\ n_{FeCl_2}=1.5L*0.25\frac{molHCl}{L} *\frac{1molHCl}{36.46gHCl}*\frac{1molFeCl_2}{2molHCl}=0.00514molFeCl_2](https://tex.z-dn.net/?f=n_%7BFeCl_2%7D%3D2.45gFe%2A%5Cfrac%7B1molFe%7D%7B55.845gFe%7D%2A%5Cfrac%7B1molFeCl_2%7D%7B1molFe%7D%3D0.0439molFeCl_2%5C%5C%5C%5C%20%20n_%7BFeCl_2%7D%3D1.5L%2A0.25%5Cfrac%7BmolHCl%7D%7BL%7D%20%2A%5Cfrac%7B1molHCl%7D%7B36.46gHCl%7D%2A%5Cfrac%7B1molFeCl_2%7D%7B2molHCl%7D%3D0.00514molFeCl_2)
In such a way, we infer the maximum moles of FeCl2 product are yielded by HCl, for which it is the limiting reactant. Finally, we calculate the grams of product by using its molar mass as shown below:
![m_{FeCl_2}=0.00514molFeCl_2*\frac{126.75gFeCl_2}{1molFeCl_2} \\\\m_{FeCl_2}=0.652gFeCl_2](https://tex.z-dn.net/?f=m_%7BFeCl_2%7D%3D0.00514molFeCl_2%2A%5Cfrac%7B126.75gFeCl_2%7D%7B1molFeCl_2%7D%20%5C%5C%5C%5Cm_%7BFeCl_2%7D%3D0.652gFeCl_2)
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Because if you have a liquid then you need a glass to keep it together and when it is a solid it is already together so you don't need to do anything
Answer:
d. In a stable molecule having an even number of electrons, all electrons must be paired.
Explanation:
<em>Which of the following statements relating to molecular orbital (MO) theory is incorrect?</em>
<em>a. Combination of two 2p orbitals may result in either σ or π MOs.</em> TRUE. If the atomic orbitals overlap frontally, a σ MO is produced while if the atomic orbitals overlap laterally, a π MO is produced.
<em>b. Combination of two atomic orbitals produces one bonding and one antibonding MO.</em> TRUE. The bonding MO has less energy than the antibonding MO.
<em>c. A bonding MO is lower in energy than the two atomic orbitals from which it is formed.</em> TRUE. The bonding MO is more stable than the individual atomic orbitals.
<em>d. In a stable molecule having an even number of electrons, all electrons must be paired.</em> FALSE. Let's consider, for example, the molecular orbitals of O₂: ![\sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2pz}^{2} \pi_{2px}^{2} \pi_{2py}^{2} \pi_{2px}^{*1} \pi_{2py}^{*1}\sigma_{2pz}^{*}](https://tex.z-dn.net/?f=%5Csigma_%7B2s%7D%5E%7B2%7D%20%5Csigma_%7B2s%7D%5E%7B%2A2%7D%20%5Csigma_%7B2pz%7D%5E%7B2%7D%20%5Cpi_%7B2px%7D%5E%7B2%7D%20%20%5Cpi_%7B2py%7D%5E%7B2%7D%20%5Cpi_%7B2px%7D%5E%7B%2A1%7D%20%5Cpi_%7B2py%7D%5E%7B%2A1%7D%5Csigma_%7B2pz%7D%5E%7B%2A%7D)
The electrons in the degenerated π* orbitals are not paired.
<em>e. A species with a bond order of zero will not be stable.</em> TRUE. For a species to be stable, the bond order must be higher than zero.
Answer:
A. Potassium phosphate: K3PO4
B. Copper (II) sulfate: CuSO4
C. Calcium chloride: CaCl2
D. Titanium dioxide: TiO2
E. Ammonium nitrate: NH4NO3
F. Sodium bisulfate: NaHSO4
Explanation:
All are ionic salts with different valent (mono valent cation or poly valent cation and anions) ions.