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3 years ago

What is the bond formation of N and H

2 answers:

8 0

Hydrogen bonding between a water molecule and an ammonia (NH3) molecule. Note that the N atom in the NH3 molecule is attracted to H atom in the H20 molecule

Zielflug [23.3K]3 years ago

3 0

Answer:

A single bond, but it's also a hydrogen bond.

Explanation:

A hydrogen bond forms when hydrogen bonds to the 3 most electronegative atoms: Oxygen, Nitrogen, or Fluorine.

It's a single bond because hydrogen only has two valence electrons. Therefore, the rest of the electrons are placed around the nitrogen atom.

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The Concentration of C6H12O6 May be represented as?

Amiraneli [1.4K]

Answer:

A 12 oz Coca Cola contains 39g of sugar or C6H12O6.

To calculate for the molarity of sugar in the soda, convert 39 grams of sugar to moles sugar:

39g/ 180.16 g/mol = 0.216 mol sugar

then, convert 12 oz to L:

12oz / (1oz/0.02957L) = 0.35484 L

therefore the concentration of sugar in the soda is:

M = mol sugar / L sol'n

= 0.216 mol sugar / 0.35484 L

= 0.609 M

Explanation:

6 0

3 years ago

Read 2 more answers

Carbon 14 decays to Carbon 12 and has a half-life of 5730 years. If a fossil is analysed and it has 5 grams of Carbon 14 and 15

Katen [24]

Answer:

im sorry but this question doesnt make sense

Explanation:

7 0

2 years ago

What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?

jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

Moles of Aluminium is 1.40 mol
Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

= 1.4 moles × 0.5

= 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

= 71.372 g

3 0

3 years ago

Convert 23.92inHg to Pa.

andriy [413]

Answer:

3189.07Pa

Explanation:

The conversion of 23.92mmH to Pa can be achieved in the following way:

760mmHg = 101325Pa

23.92mmHg = (23.92x101325)/760 = 3189.07Pa

5 0

3 years ago

Read 2 more answers

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

Now put all the given values in this formula, we get:

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago