Answer:
A) = 4.7 × 10⁻⁴atm
Explanation:
Given that,
Kp = 1.5*10³ at 400°C
partial pressure pN2 = 0.10 atm
partial pressure pH2 = 0.15 atm
To determine:
Partial pressure pNH3 at equilibrium
The decomposition reaction is:-
2NH3(g) ↔N2(g) + 3H2(g)
Kp = [pH2]³[pN2]/[pNH3]²
pNH3 =√ [(pH2)³(pN2)/Kp]
pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm
![K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BpH_2%5D%20%5E3%5BpN_2%5D%7D%7B%5BpNH_3%5D%5E2%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%28pH_2%29%5E3%28pN_2%29%7D%7BpNH_3%7D%20%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.15%29%5E3%280.10%29%7D%7B1.5%20%5Ctimes%2010%5E3%7D%20%7D%20%5C%5C%3D4.74%20%5Ctimes%2010%5E-%5E4atm)
= 4.7 × 10⁻⁴atm
When an ionic is placed in water a dissolving reaction occurs so the positive or negative ion are only attracted to each other
1. Deeply embedded pearls.
2. All of the above
3. Fluid trapped in sedimentary
4. All of the above
Answer:
![Molec_{\ H_{tot}}=1.206x10^{25}molec](https://tex.z-dn.net/?f=Molec_%7B%5C%20H_%7Btot%7D%7D%3D1.206x10%5E%7B25%7Dmolec)
Explanation:
Hello.
In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:
![molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH} =6.03x10^{22}molec](https://tex.z-dn.net/?f=molec_%7B%5C%20H%7D%3D3.65gHCl%2A%5Cfrac%7B1molHCl%7D%7B36.45gHCl%7D%20%2A%5Cfrac%7B1molH%7D%7B1molHCl%7D%2A%5Cfrac%7B6.022x10%5E%7B23%7Dmolec_%5C%20H%7D%7B1molH%7D%20%20%3D6.03x10%5E%7B22%7Dmolec)
Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:
![molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH} =1.20x10^{25}molec](https://tex.z-dn.net/?f=molec_%7B%5C%20H%7D%3D180gH_2O%2A%5Cfrac%7B1molH_2O%7D%7B18gH_2O%7D%20%2A%5Cfrac%7B2molH%7D%7B1molH_2O%7D%2A%5Cfrac%7B6.022x10%5E%7B23%7Dmolec_%5C%20H%7D%7B1molH%7D%20%20%3D1.20x10%5E%7B25%7Dmolec)
Thus, the total number of molecules turns out:
![Molec_{\ H_{tot}}=6.03x10^{22}+1.20x10^{25}\\\\Molec_{\ H_{tot}}=1.206x10^{25}molec](https://tex.z-dn.net/?f=Molec_%7B%5C%20H_%7Btot%7D%7D%3D6.03x10%5E%7B22%7D%2B1.20x10%5E%7B25%7D%5C%5C%5C%5CMolec_%7B%5C%20H_%7Btot%7D%7D%3D1.206x10%5E%7B25%7Dmolec)
Regards.