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Svetllana [295]
3 years ago
9

certain c-h groups can form weak hydrogen bonds. why would such a group be more likely to be a hydrogen bond donor group when th

e c is next to n
Chemistry
1 answer:
damaskus [11]3 years ago
5 0

Answer:

C-H groups can form weak hydrogen bonds and such compounds are more likely to have hydrogen bond donor groups because the hydrogen group is highly polar which is bonded to a strongly electronegative atom such as carbon, nitrogen and oxygen.

As the atom bonded with hydrogen atom have an equivalent partial negative charge, so it can only accept H-bonds from other atoms.

Thus, c-h groups forming weak hydrogen bonds are more likely to be hydrogen bond donor group.

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Calculate the molality of a 0.677 m ethanol ( c 2 h 5 o h ) solution whose density is 0.588 g/ml
ANEK [815]

The molality of a 0.677 m ethanol ( c 2 h 5 o h ) solution whose density is 0.588 g/ml.

What is molality and molarity?

The number of solvent moles per kilogram is known as molality. Because the mass of the solute and solvent in the solution remains constant, molality is the preferred concentration transfer method.

Molarity, also referred to as molar concentration, is the quantity of a material expressed as moles per litre of solution. Solutions marked with a molar concentration have a capital M. One mole of solvent per litre is present in a 1.0 M solution.

Molar mass (w) of C2H5OH = 46g

Molarity(M) = 0.677M

Density(D) = 0.588g/mol

<u>Molality(m)</u><u> = M/D * (1000 + w*M)</u>

<u>= 1187.2m</u>

To learn more about molality and molarity  from the given link below,

brainly.com/question/14770448

#SPJ4

4 0
1 year ago
A compound contains 24.27% C, 4.07% H and 71.65% Cl. The molar mass is about 99g.
Schach [20]

Answer:

Molecular formula = C₂H₄Cl₂

Explanation:

Molecular formula:

Molecular formula consist of symbols of elements present in compound with numbers of atoms of each element in subscript.

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number .

Given data:

Percentage of hydrogen = 4.07%

Percentage of Cl = 71.65%

Percentage of carbon = 24.27%

Molar mass = 99 g/mol

Molecular formula = ?

Solution:

Number of gram atoms of H = 4.07 / 1.01 = 4.0

Number of gram atoms of Cl = 71.65 / 35.5 = 2.0

Number of gram atoms of C = 24.27 / 12 = 2.0

Atomic ratio:

           H              :          C             :         Cl

          4/2            :        2/2            :       2/2

           2               :          1             :        1

C : H : Cl = 1 : 2 : 1

Empirical formula is CH₂Cl.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass  = 12+1×2+ 35.5 = 49.5

n =  99/49.5

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 ( CH₂Cl)

Molecular formula = C₂H₄Cl₂

5 0
3 years ago
Which is used to convert between mass and moles of a substance?
sveticcg [70]
The molar mass constant is used to convert mass to moles 
3 0
3 years ago
___ V2O5 + ___ CaS ___ CaO + ___ V2S5
den301095 [7]

Answer:

3

Explanation:

5 0
3 years ago
Determine the chemical formula for the compound, tetracarbonylplatinum(iv) chloride. [pt(co)4cl4] 2- [ptcl4](co)4 [pt(co)4]cl4 [
shusha [124]
Analyze the Name of complex Compound.

                              T<span>etracarbonylplatinum(iv) chloride

So, there are,
                               4 Carbonyl groups = 4 CO = (CO)</span>₄
                               1 Platinum Metal    = 1 Pt  = Pt
                               Unknown Chloride atoms = ?
In complexes positive part is always named first, so the sphere containing Pt and carbonyl ligands is written first,

                                     [Pt (CO)₄]

The charge on sphere is +4 because CO ligand is neutral, and Pt has a Oxidation state of four as written in name (IV),
So,
                                             [Pt (CO)₄]⁴⁺

Now, in order to neutralize +4 charge we should add 4 Chloride ions, So,

                                            [Pt (CO)₄] Cl₄
4 0
3 years ago
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