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Feliz [49]
3 years ago
8

Compare the dipole moments of the following pairs of molecules and determine whether the first is "Greater Than", "Less Than" or

"Equal to" the second.
O The dipole moment of ClO2 is _____ the dipole moment of SO2
O The dipole moment of SiF4 is _____ the dipole moment of SF4
O Dipole moment of SO2 is ______the dipole moment of SO3
O The dipole moment of BeCl2 is _____ the dipole moment of SCl2
O The dipole moment of H2O is ______ the dipole moment of H2S
Chemistry
1 answer:
castortr0y [4]3 years ago
6 0

Answer:

Greater than

Less than

Greater than

Less than

Greater than

Explanation:

The dipole moment is the difference of electronegativity between the atoms in a molecule, and the total is the sum of the dipole moments of the bonds. So, it depends on the geometry of the molecule. When the dipole moment is 0 the molecule is nonpolar, when it is different from 0 the molecule is polar.

Both ClO₂ and SO₂ have angular geometry because there are lone pairs of electrons in the central atom (Cl and S), but Cl has a higher value of electronegativity than S, so the dipole moment must be greater.

SiF₄ is a nonpolar molecule, which has tetrahedral geometry and no lone pairs at the central atom. SF₄ has lone pairs at the central atom, and then the molecule is polar, so the dipole moment of SiF₄ is less than of SF₄.

The SO₃ molecule has no lone pairs at the central atom and has trigonal geometry, so it's a nonpolar molecule. SO₂ has angular geometry and it's a polar molecule, so the dipole moment of SO₂ is greater.

BeCl₂ has a linear geometry and is a nonpolar molecule. SCl₂ has an angular geometry, and it's a polar molecule, so the dipole moment of BeCl₂ is less than the dipole moment of SCl₂.

Oxygen has a higher electronegativity than the sulfur, and both molecules are polar with angular geometry, so the dipole moment of H₂O is greater.

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never [62]

creo esto te va a ayudar

5 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
This question is about metal oxides. When sodium is heated in oxygen, sodium oxide is produced.
vodomira [7]

Answer:

This is very easy Cuz We have 2Na2O We have O2 so thats molecule of Oxygen and its same on product We need to balance Na on start We have 1 on product We have 2 so Just put 2 at start....

Explanation:

Sorry for bad english not my first language :(

2Na+O2-->2Na20

5 0
3 years ago
EL hidroxido de sodio reacciona con el sulfato de hierro (II) para formar sulfato de sdio e hidroxido ferroso. Si se hace reacci
Oduvanchick [21]

Answer:

280.8 g

Explanation:

Definimos la reaccion:

2NaOH +  FeSO₄  →  Na₂SO₄  +  Fe(OH)₂

Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.

Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol

2 moles de NaOH producen 1 mol de hidroxido ferroso

Entonces 6.25 moles producirán, la mitad (6.25  . 1) /2 = 3.125 moles

Convertimos los moles a masa:

3.125 mol . 89.85 g/mol = 280.8 g

7 0
3 years ago
Plsss help i’ll give brainlist
yarga [219]

Answer:

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Explanation:

7 0
2 years ago
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