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DochEvi [55]
3 years ago
10

Urea is an organic compound widely used as a fertilizer. Its solubility in water allows it to be made into aqueous fertilizer so

lutions and applied to crops in a spray. What is the maximum theoretical number of water molecules that one urea molecule can hydrogen bond with?
Chemistry
1 answer:
meriva3 years ago
8 0

Answer:

8 water molecules

Explanation:

The hydrogen bond may be H-O~H-N or  H-N~H-O; in the first one, the hydrogen bond is between an oxygen atom and a hydrogen which is covalently bonded to a nitrogen atom. The second one is the hydrogen bond of a nitrogen atom with a hydrogen covalently bonded to a oxygen one. The first case would be the hydrogen bonds that water may form with the hydrogen of the urea; the second ones would be the hydrogen bonds that urea may form with water molecules. So, for each nitrogen in urea there would be a hydrogen bond, and for each hydrogen too. Finally, the oxygen in the urea molecule may form hydrogen bonds with water as well, but it has two lone pairs to donate, so the oxygen atom may form hydrogen bond with 2 water molecules:

N=(2 because of the oxygen atom of the urea)+(4 because of the hydrogen bonded to nitrogen)+2(because of the nitrogens).

N=8.

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An antifreeze solution is made by mixing ethylene glycol (ρ = 1116 kg/m3) with water. suppose the specific gravity of such a sol
Harman [31]
Answer is: <span>the volume percentage of ethylene glycol in the solution is 43.1%.
V(solution) = 1 m</span>³.
d(solution) = 1050 kg/m³.
m(solution) = 1050 kg.
d(ethylene glycol) = 1116 kg/m³.
d(water) = 1000 kg/m³.
m(solution) = d(ethylene glycol) · V(ethylene glycol) + d(water) · V(water). 
V(ethylene glycol) = 1m³ - V(water).
1050 kg = 1116 kg/m³ · (1m³ - V(water)) + 1000 kg/m³ · V(water).
1050 kg = 1116 kg - 1116·V(water) + 1000·V(water).
116 kg/m³ ·V(water) = 66kg.
V(water) = 0.569 m³.
V(ethylene glycol) = 1 m³ - 0.569 m³ = 0.431 m³.
percentage = 0.431 m³ ÷ 1 m³ · 100% = 43.1%.

3 0
3 years ago
2 What has been one effect of population growth on the physical environment? ​
miskamm [114]

Answer:

It damaged the environment growth of humans and the population had a negative affect on the environment

Explanation:

8 0
2 years ago
How would you prepare 250 mL of a 0.100 M solution of fluoride ions<br> from solid CaF2?
andrey2020 [161]

In 250 mL of volumetric flask add 0.975875 grams of CaF_2 and dissolve it in the 250 mL of water.

Given:

The solid of calcium fluoride.

To prepare:

The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.

Method:

Molarity of the fluoride ion solution needed = M = 0.100 M

The volume of the fluoride ion solution needed = V = 250 mL

1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L

The moles of fluoride ion needed = n

According to the definition of molarity:

M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol

Moles of fluoride ion = 0.025 mol

We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.

CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)

According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:

=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2

Moles of calcium fluoride = 0.0125 mol

Mass of calcium fluoride needed to prepare the solution :

=0.0125 mol\times 78.07 g/mol=0.975875 g

Preparation:

  • Weight 0.975875 grams of calcium fluoride
  • Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
  • Now add a small amount of water to dissolve the calcium fluoride completely.
  • After this add more water up to the mark of the volumetric flask of volume 250 mL.

Learn more about molarity of solution ere:

brainly.com/question/10053901?referrer=searchResults

brainly.com/question/10270173?referrer=searchResults

3 0
3 years ago
Greg classified some substances as shown in the table. Greg's Table Soluble in Water Insoluble in Water Dishwashing detergent Cl
densk [106]

Greg’s table is incorrect because sand is insoluble in water.

6 0
2 years ago
Read 2 more answers
Considering 2N2H4(g) + N2O4(g) -&gt; 3N2(g) +4H2O(g)
Tema [17]

Answer:

107.8

Explanation:

64 gram of N2H4 produce 72 gram of H20

then by crossmultiplication

64*121.3/72=107.82

3 0
2 years ago
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