Answer:
The molecular formula of the compound is
. The molecular formula is obtained by the following expression shown below
![\textrm{Molecular formula }= n\times \textrm{Empirical formula}](https://tex.z-dn.net/?f=%5Ctextrm%7BMolecular%20formula%20%7D%3D%20n%5Ctimes%20%5Ctextrm%7BEmpirical%20formula%7D)
Explanation:
Given molecular mass of the compound is 176 g/mol
Given empirical formula is
Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.
Empirical formula mass of the compound = ![\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}](https://tex.z-dn.net/?f=%5Cleft%20%28%202%5Ctimes12%2B4%2B16%20%5Cright%20%29%20%5Ctextrm%7B%20u%7D%20%3D%2044%20%5Ctextrm%7B%20g%2Fmol%7D)
![n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4](https://tex.z-dn.net/?f=n%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5Ctextrm%7BMolecular%20formula%20mass%7D%7D%7B%5Ctextrm%7BEmpirical%20formula%20mass%7D%7D%20%5C%5Cn%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B176%7D%7B44%7D%20%3D%204)
![\textrm{Molecular formula }= n\times \textrm{Empirical formula}](https://tex.z-dn.net/?f=%5Ctextrm%7BMolecular%20formula%20%7D%3D%20n%5Ctimes%20%5Ctextrm%7BEmpirical%20formula%7D)
Molecular formula = 4
Molecular formula is ![C_{8}H_{16}O_{4}](https://tex.z-dn.net/?f=C_%7B8%7DH_%7B16%7DO_%7B4%7D)
London Dispersion forces
Explanation:
London dispersion forces exists between the molecules of CO₂. These are weak attractions found between non-polar (and polar) molecules(symmetrical molecules).
- This attraction sometimes makes non-polar molecules polar.
- This is because of the uneven charge distribution caused by the constant motion of its electrons.
- A temporary dipole or instantaneous dipole induces the neighboring molecules to become distorted and forms dipoles as well.
- The forces are the weakest of all electrical forces.
Learn more:
Intermolecular forces brainly.com/question/10107765
#learnwithBrainly
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:
![\% M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\% =\frac{m_M}{m_M+32.00}*100 \%](https://tex.z-dn.net/?f=%5C%25%20%20M%3D%5Cfrac%7Bm_M%7D%7Bm_M%2B2%2Am_O%7D%2A100%20%5C%25%5C%5C%5C%5C59.93%5C%25%20%20%3D%5Cfrac%7Bm_M%7D%7Bm_M%2B32.00%7D%2A100%20%5C%25)
Thus, we solve for the molar mass of the metal to obtain:
![59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol](https://tex.z-dn.net/?f=59.93%5C%25%20%28m_M%2B32.00%29%20%3Dm_M%2A100%20%5C%25%5C%5C%5C%5Cm_M%2A59.93%5C%25%20%2B1917.76%5C%25%20%3Dm_M%2A100%20%5C%25%5C%5C%5C%5Cm_M%3D47.86g%2Fmol)
For the subsequent problems, we proceed as follows:
a.
![4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2](https://tex.z-dn.net/?f=4.00gO_2%2A%5Cfrac%7B1molO_2%7D%7B32.00gO_2%7D%3D0.125molO_2)
b.
![0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH](https://tex.z-dn.net/?f=0.400molH_2S%2A%5Cfrac%7B2molH%7D%7B1molH_2S%7D%2A%5Cfrac%7B6.022x10%5E%7B23%7DatomsH%7D%7B1molH%7D%3D4.82x10%5E%7B23%7DatomsH)
c.
![0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH](https://tex.z-dn.net/?f=0.235gNH_3%2A%5Cfrac%7B1molNH_3%7D%7B17.03gNH_3%7D%20%2A%5Cfrac%7B3molH%7D%7B1molNH_3%7D%2A%5Cfrac%7B6.022x10%5E%7B23%7DatomsH%7D%7B1molH%7D%3D2.49x10%5E%7B22%7DatomsH)
Regards!
Titanium has an atomic number of 22, which means it has 22 protons and 22 electrons. About 73.8 percent of natural titanium exists as the isotope Ti-48, which contains 26 neutrons.