Answer:
Covalent bonds.
Explanation:
Covalent bonds are intermolecular bonds which involve the sharing of electrons between atoms or molecules and ions with close or near electronegativity values.
This usually form molecules. This is also the sharing of electrons donated by one of the atoms or by both of the atoms to form a molecule and at the same time obtain a stable octet state.
We need to know the number of molecules which is number of moles * Avogadro’s number
So
We need to compare number of moles because Avogadro’s numbers is constant for all
We’ll start with Nitrogen gas : 14g/28g.mol=0.5 mole
For NH3 : 3.4g / (14+3)g.mol= 0.2 mol
For NO. : 15g/(14+16)g.mol = 0.5 mol
For O2 : 64g/32 g.mol = 2 mol
For SO2: 32g /( 32+32) g.mol = 0.5mol
For H2. : 1g/ 2g.mol = 0.5 mol
So
H2& SO2 & NO have the same number of moles of N2 and the same number of molecules of N2
Answer:
200cc of 5N HCl were diluted
Explanation:n
<em>Assuming the solution was diluted to prepare a 1N HCl solution:</em>
The HCl was diluted from 5N to 1N, that is a dilution of:
5N / 1N = 5 times
As the total volume of the diluted solution is 1L = 1000mL = 1000cc the amount of the concentrated solution was:
1000cc / 5 =
<h3>200cc of 5N HCl were diluted</h3>
Answer:
![m_{Fe_3O_4}=1.7gFe_3O_4](https://tex.z-dn.net/?f=m_%7BFe_3O_4%7D%3D1.7gFe_3O_4)
Explanation:
Hello,
In this case, considering the given reaction:
![3 Fe(s) + 4 H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4 H_2(g)](https://tex.z-dn.net/?f=3%20Fe%28s%29%20%2B%204%20H_2O%28g%29%20%5Crightleftharpoons%20Fe_3O_4%28s%29%20%2B%204%20H_2%28g%29)
Thus, for the equilibrium, just water and hydrogen participate as iron and iron(II,III) oxide are solid:
![Kc=\frac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Thus, at the beginning, the concentration of water is 0.05 M and consequently, at equilibrium, considering the ICE procedure, we have:
![5.1=\frac{(4x)^4}{(0.05M-4x)^4}](https://tex.z-dn.net/?f=5.1%3D%5Cfrac%7B%284x%29%5E4%7D%7B%280.05M-4x%29%5E4%7D)
Thus, the change
is obtained as:
![\sqrt[4]{5.1} =\sqrt[4]{[\frac{(4x)}{(0.05M-4x)}]^4}\\\\1.5=\frac{(4x)}{(0.05M-4x)}\\\\x=0.0075M](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5.1%7D%20%3D%5Csqrt%5B4%5D%7B%5B%5Cfrac%7B%284x%29%7D%7B%280.05M-4x%29%7D%5D%5E4%7D%5C%5C%5C%5C1.5%3D%5Cfrac%7B%284x%29%7D%7B%280.05M-4x%29%7D%5C%5C%5C%5Cx%3D0.0075M)
Thus, the moles of hydrogen at equilibrium are:
![[H_2]_{eq}=4*0.0075\frac{mol}{L}*1.0L=0.03molH_2](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D4%2A0.0075%5Cfrac%7Bmol%7D%7BL%7D%2A1.0L%3D0.03molH_2)
Therefore, the grams of iron(II,III) oxide finally result:
![m_{Fe_3O_4}=0.03molH_2*\frac{1molFe_3O_4}{4molH_2}*\frac{231.533gFe_3O_4}{1molFe_3O_4} \\m_{Fe_3O_4}=1.7gFe_3O_4](https://tex.z-dn.net/?f=m_%7BFe_3O_4%7D%3D0.03molH_2%2A%5Cfrac%7B1molFe_3O_4%7D%7B4molH_2%7D%2A%5Cfrac%7B231.533gFe_3O_4%7D%7B1molFe_3O_4%7D%20%5C%5Cm_%7BFe_3O_4%7D%3D1.7gFe_3O_4)
Best regards.