Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
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Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
Whereas the moles of the salt are computed as shown below:
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
![[salt]=0.01428mol/0.0276L=0.517M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D0.01428mol%2F0.0276L%3D0.517M)
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
Whose equilibrium expression is:
![Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5NH_2%5D%5BH_3O%5E%2B%5D%7D%7BC_6H_5NH_3%5E%2B%7D)
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
Whereas x is:
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
Regards!
Answer:
MnO2 M n O 2 is an ionic substance without any oxo-anions, therefore its name has the -ide ending.
<span>If you do not wash and dry the thermometer after every time you use it in this scenario, the temperature could be affected by the NaOH residue. This could make it so that your findings were inaccurate, so in order to be efficient, this precaution needs to be taken.</span>
