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stira [4]
3 years ago
14

In two or more complete sentences describe all of the van der Waals forces that exist between molecules of sulfur

Chemistry
2 answers:
Pani-rosa [81]3 years ago
8 0

Answer:

The correct answer is "there are many Van der Waals forces acting between molecules of sulfur dioxide"

Explanation:

Van Der Waals forces are considered one of the weakest forces, however, since they exist in multiple levels, they have a major influence in molecules behavior. Van Der Waals forces are intermolecular interactions that emerge from the electrons movement in the atoms and the place that the electrons have in a given moment. In the case of sulfur dioxide, the atoms of sulfur and oxygen have Van Der Waals forces within them and among them. Particularly, the electrons within the individual atoms, as well as the electrons able to interact among the sulfur dioxide molecules, have all Van Der Waals forces.

zavuch27 [327]3 years ago
3 0

Answer:

Dipole-Dipole attraction

Explanation:

Dipole-dipole attraction is a type of vander waals forces found in the molecules of sulfur dioxide.

Vander waals forces are weak attractions joining non-polar and polar molecules together. They are of two types:

  • London dispersion forces which are weak attractions found between non-polar molecules.
  • Dipole-Dipole attraction are the forces of attraction which exists between polar molecules. Such molecules have permanent dipoles. This implies that the positive pole of one molecule attracts the negative pole of another. This is what happens between the oxygen and sulfur molecules.
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Answer:

the Germany scientist Alfred Wagner describes fold mountains that are found in The world are the same also geological similarity that occur between this continent are the same for example the rock that are found in brazil are the same of that of south africa.

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Sodium bromate chemical formula
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A sample of gas occupies 10.0 l at 100.0 torr and 27.0
disa [49]
The pressure  of a sample  of a gas if the temperature  is changed  to 127  c  while  the volume  remains  constant   is calculated  using gay lussac law formula

that is P1/T1 = P2/V2
P1 = 100 torr
T1 = 27+273 = 300 k
T2 =127 +273 =400 k
P2=?

by  making P2 the subject of the formula
P2=T2P1/T1

=100  x 400/300 = 133.3 torr

5 0
4 years ago
you fill a rigid steel container that has a volume of 20 L with nitrogen gas to a final pressure of 2 x 10^4 kpa at 23 Celsius.
garik1379 [7]

Answer:

4.549 kg.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 2 x 10⁴ kPa/101.325 = 197.4 atm).

V is the volume of the gas in L (V = 20.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 23° C + 273 = 296 K).

<em>∴ n = PV/RT =</em> (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = <em>162.5 mol.</em>

  • To find the mass of N₂ in the cylinder, we can use the relation:

<em>mass of N₂ = (no. of moles of N₂)*(molar mass of N₂) = </em>(162.5 mol)*(28.0 g/mol) = <em>4549 g = 4.549 kg.</em>

3 0
3 years ago
Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to
Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density \rho_{avg}; we have:

\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }

The average atomic weight is:

A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

C_{Fe} = 100 - C_v

From a unit cell volume V_c

V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}

where;

N_A = number of Avogadro constant.

SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

C_{Fe} with 100-C_v

Then:

a^3 = \dfrac   { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) }    {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

Replacing the values; we have:

(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol)   }   }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3)   } }

2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} }  \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }

2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})

\mathbf{C_v = 9.1 \ wt\%}

4 0
3 years ago
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