Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
Answer:
the double bond between c and o is shorter and weaker
Explanation:
this is because the bond between c and o involves unequal sharing of electrons whole c and c involves hybridization sp2 of orbitals and also catenation phenomenon in which carbon could form long chain with it's other carbon
Answer:
Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.
Explanation:
Hope this helped :)
Answer:
Explanation:
Burning fossil fuels releases the carbon dioxide stored millions of years ago. ... The concentration of carbon dioxide in the atmosphere has increased more in the northern hemisphere where more fossil fuel burning occurs. Since the Industrial Revolution the concentration globally has increased by about 40 % .
Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.
=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 
The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
