You have molarity and you have volume. Use the formula :
Molarity(M)= Moles(N)/Liter(L) to get the solution.
150 ml= .150 L
7.7 = N/.150
N=.1.155 moles of NaOH.
And since you know the moles, use the molar mass to figure out the grams.
<span> (40g/mol NaOH) x (1.155mol) =
46.2 g of NaOH.</span>
1) Carbon-13:
Proton-6 Neutron-7 Electron-6
2)Atomic mass of element X:
(55*10+56*20+57*70)/100=56.6
Answer:
%
Explanation:
The ethanol combustion reaction is:
→![2CO_{2}+3H_{2}O](https://tex.z-dn.net/?f=2CO_%7B2%7D%2B3H_%7B2%7DO)
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
![x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}](https://tex.z-dn.net/?f=x%2A1.10%28excess%29%2A%5Cfrac%7B3%20O_%7B2%7Dmoles%20%7D%7BetOHmole%7D)
Dividing the previous equation by x:
![1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}](https://tex.z-dn.net/?f=1.10%28excess%29%2A%5Cfrac%7B3%20O_%7B2%7Dmoles%7D%7BetOHmole%7D%3D3.30%5Cfrac%7BO_%7B2%7Dmoles%7D%7BetOHmole%7D)
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
![3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )](https://tex.z-dn.net/?f=3.30%28O_%7B2%7D%20moles%29%2A%5Cfrac%7B0.79%28N_%7B2%7D%20moles%29%7D%7B0.21%28O_%7B2%7D%20moles%29%7D%3D121.414%20%28N_%7B2%7D%20moles%20%29)
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
![3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles](https://tex.z-dn.net/?f=3.30%28O_%7B2%7D%20moles%29-0.85%28etOHmoles%29%2A%5Cfrac%7B3%28O_%7B2%7D%20moles%29%7D%7B1%28etOHmoles%29%7D%20%3D0.75O_%7B2%7D%20moles)
Calculate the number of moles of CO2 and water considering the same:
![0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)](https://tex.z-dn.net/?f=0.85%28etOHmoles%29%2A%5Cfrac%7B3%28H_%7B2%7DOmoles%29%7D%7B1%28etOHmoles%29%7D%3D2.55%28H_%7B2%7DOmoles%29)
![0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)](https://tex.z-dn.net/?f=0.85%28etOHmoles%29%2A%5Cfrac%7B2%28CO_%7B2%7Dmoles%29%7D%7B1%28etOHmoles%29%7D%3D1.7%28CO_%7B2%7Dmoles%29)
The total number of moles at the reactor output would be:
![N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)](https://tex.z-dn.net/?f=N%3D1.7%28CO2%29%2B12.414%28N2%29%2B2.55%28H2O%29%2B0.75%28O2%29%5C%5C%20N%3D17.414%28Dry-air-moles%29)
So, the oxygen mole fraction would be:
%
Answer:
![\boxed{\text{14}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B14%7D%7D)
Explanation:
If l = 3, the electrons are in an f subshell.
The number of orbitals with a quantum number l is 2l + 1, so there
are 2×3 + 1 = 7 f orbitals.
Each orbital can hold two electrons, so the f subshell can hold 14 electrons.
![\boxed{\textbf{14 electrons}} \text{ can share the quantum numbers n = 4 and l = 3.}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctextbf%7B14%20electrons%7D%7D%20%5Ctext%7B%20can%20share%20the%20quantum%20numbers%20n%20%3D%204%20and%20l%20%3D%203.%7D)