The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).
Answer:
Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.
hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]
So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.
If one isomer of the alkene is trans then the other two isomers may be cis .
Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.
The two possibility of cis structures are possible:
in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.
Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.
Kindly refer the attachments for the structure of compounds:
Answer:
Al 72.61%
Mg 27.39%
Explanation:
To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.
If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.
To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.
Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g
The mass of magnesium is 0.0381 * 24 = 0.9144g
We can now calculate the mass percentage.
The total mass is 0.9144 + 2.4246 = 3.339g
% mass of Al = 2.4246/3.339 * 100 = 72.61%
% mass of Mg = 0.9144/3.39 * 100 = 27.39%
Answer:p-hydroxybenzaldehyde is stronger acid to phenol
para-cyanophenol is stronger acid to meta-cyanophenol
o-fluorophenol is stronger acid to p-fluorophenol.
Explanation:
The PKa tool relative to Ph are used to contrast the pairs.
The pKa of phenol is 10. The pKa of p-hydroxybenzaldehyde is 9.24
The pKa for meta-cyanophenol is 8.61 and the pKa for para-cyanophenol is 7.95.
The pKa value of o-fluorophenol is 8.7, while that of the p-fluorophenol is 9.9. It's obvious that the inductive effect is more dominant at ortho-position, which results in a more acidic nature
The pKa is the pH value at which a chemical species will accept or donate a proton. The lower the pKa, the stronger the acid and the greater the ability to donate a proton in aqueous solution.
The hydroxide concentration is written as [OH-]. A useful equation is that pH + pOH = 14. Since the pH is 4, the pOH must be 10. To get [OH-], again take the neg. anti-log of both sides: - anti log (pOH-) = - antilog(10) ---> [OH-]=1 x 10^-10 M or 1 x 10^-10 moles [OH-] per liter solution (in this case, lake).
The hydroxide ion concentration describe in question 3 is 10-10 M
