Equation of bhr and h2o

This is a specific type of dipole-dipole force that exists between molecules with hydrogen atoms and molecules with nitrogen, ox

RoseWind [281]

Answer: This description refers to: "hydrogen bonding" .
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 Hydrogen bonding is a specific type of "dipole-dipole force" that exists between [molecules with hydrogen atoms] and [molecules with nitrogen, oxygen, or fluorine atoms.] .
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5 0

2 years ago

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What is the mole fraction of cation in 1M of aluminium sulphate

Svetllana [295]

The mole fraction of cation (Al)=0.4

<h3>Further explanation</h3>

Given

1 M of Aluminium sulphate

Required

The mole fraction of cation

Solution

Ionization of the Aluminum sulfate solution( assume 1 L solution ) :

mol Al₂(SO₄)₃ = M x V = 1 M x 1 L = 1 mol

Al₂(SO₄)₃⇒2Al³⁺ + 3SO₄²⁻

1 mol 2 mol 3 mol

From this equation, total mol in solution = 2+3 = 5 moles

Mol fraction Al(as a cation) :

= 2/5=0.4

8 0

3 years ago

Which hybrid orbitals overlap to form the sigma bonds between the indicated atoms in xanthine?

amid [387]

Answer:

C.) Csp² - Nsp³

Explanation:

Because the carbon has only 3 unique bonds, the conformation of the carbon should be Csp² (s + p + p). While the nitrogen also has only 3 bonds, making up 3 orbitals, it also has a lone pair of electrons. These electrons also take up one orbital. Therefore, the conformation of nitrogen is Nsp³ (s + p + p + p).

8 0

1 year ago

In a hydrogen atom, the electron and the proton are seperated by about 0.5 angstroms, 5.0×10−11 m. what is the acceleration of a

nadya68 [22]

The charge of proton and electron is equal in magnitude that is $1.6\times 10^{-19} C$ . The distance between electron and proton is $0.5 \AA$ or $5\times 10^{-11} m$ .

Force exerted on electron by proton is calculated as follows:

$F= \left | \frac{kq_{e}q_{p}}{r^{2}} \right |$

Here, k is Coulomb's constant, $q_{p}$ is charge on proton, $q_{e}$ charge on electron and r is distance between them.

Putting the values,

$F= \left | \frac{(8.99\times 10^{9}N m^{2}/C^{2})(1.6\times 10^{-19} C)(1.6\times 10^{-19} C)}{(5\times 10^{-11} m)^{2}} \right |=9.2\times 10^{-8} N$

Now, according to second law of Newton,

$F=m_{e}a$

Here, $m_{e}$ is mass of electron and a is acceleration.

Also, $1 N=1 kgm/s^{2}$

Rearranging,

$a=\frac{F}{m_{e}}=\frac{9.2\times 10^{-8} Kg m/s^{2}}{9.11\times 10^{-31} kg}=1.01\times 10^{23} m/s^{2}$

Thus, the acceleration of an electron in this atom is $1.01\times 10^{23} m/s^{2}$ .

7 0

3 years ago

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1. Provide at least two statements that explain the relationship between the position of the Earth and the seasons.

balu736 [363]

Statement 1: The Earth's axis tilt makes one hemisphere face away from the sun during orbit, creating cooler seasons.
Statement 2: The Earth's axis tilt makes one hemisphere face towards the sun during orbit creating warmer seasons.

3 0

2 years ago