Answer: The IUPAC name of 
is 5-chloro-2-pentyne
Explanation:
1. First select the longest possible carbon chain. For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
2. The longest possible carbon chain should contain all the bonds and functional groups.
3. The numbering is done in such a way that the carbon containing the functional group or substituent gets the lowest number. Triple bond is given priority over substituent halogen.
4. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne.
Thus the IUPAC name of is 5-chloro-2-pentyne
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
Answer:
Explanation:
The pressure is constant, so we can use Charles' Law to calculate the volume.
Data:
V₁ = 693 mL; T₁ = 45 °C
V₂ = ?; T₂ = 150 °C
Calculations:
(a) Convert temperature to kelvins
T₁ = ( 45 + 273.15) = 318.15 K
T₂ = (150 + 273.15) = 423.15 K
(b) Calculate the volume
Molar mass of C6H12O6 = 6x12 + 12x1 + 6x16 = 180g
SO mass of 1 mole of C6H12O6 = 180g
mass of 0.5 mole of C6H12O6 = 0.5 x180g = 90g
