As number of gaseous moles in reactant and prodict are same that is 4
So No change will occur
Answer:
b) 0.47
Explanation:
MwC5H12 = 72.15g/mol
⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12
MwC6H14=86.18g/mol
⇒molC6H14=(20.0)*(mol/86.18)=0,232
MwC6H6=78.11g/mol
⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6
<h3>XC6H14=(0.232)/(0.1386+0.232+0,128)=0.465≅0.47</h3>
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.
Answer:
The number on the lag label should be 15.
Explanation:
It seems your question is incomplete, as it is lacking the working values. An internet search showed me the full question, you can see it in the attached picture.
Let's say we have 100 g of the fertilizer.
- <em>45 g are of ammonium phosphate</em> ( (NH₄)₃PO₄ ), of which:
- 45 g (NH₄)₃PO₄ *
= 12.7 g are of Nitrogen.
(We used the molar mass of ammonium phosphate in the denominator and three times the molar mass of nitrogen in the numerator)
- <em>18 g are of calcium nitrate</em> (Ca(NO₃)₂), of which:
- 16 g Ca(NO₃)₂ *
= 2.73 g are of Nitrogen.
So in total there are (12.7+2.73) 15.43 g of Nitrogen in 100 g of the fertilizer. So the percent by mass of nitrogen is 15.43%.
Rounding to the nearest percent the answer is 15.