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3 years ago

What causes the difference in bond angles in carbon dioxide and water?

Chemistry

2 answers:

puteri [66]3 years ago

8 0

I would say the answer is C. The number of lone pairs of electrons on the central atom is greater in the case of water.

egoroff_w [7]3 years ago

5 0

Answer is: C) the fact that the number of lone pairs of electrons on the central atom is greater in the case of water.

Carbon(IV) oxide is nonpolar because CO₂ is linear molecule and the oxygen atoms are symmetrical (bond angles 180°).

Water is polar because of the bent shape of the molecule.

Oxygen atom in water molecule has sp3 hybridization. The bond angle between the two hydrogen atoms is approximately 104.45°.

Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen atom has six valence electrons , two lone pairs and two electrons that form two sigma bonds with hydrogen atoms.

Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.

Electron configuration of carbon atom: ₆C 1s² 2s² 2p².

In carbon dioxide, carban has sp hybridization with no lone pairs.

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The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Alex_Xolod [135]

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

4 0

3 years ago

Read 2 more answers

The study of all chemicals containing carbon is called ?

Jlenok [28]

The study of all chemicals containing carbon is organic chemistry.

5 0

3 years ago

A 2.3 kg object has 15 ) of kinetic energy. Calculate its speed.

chubhunter [2.5K]

Answer:

<h3>The answer is option D</h3>

Explanation:

To find the speed given the kinetic energy and mass we use the formula

$v = \sqrt{ \frac{2KE}{m} } \\$

where

m is the mass

v is the speed

From the question

KE = 15 J

m = 2.3 kg

We have

$v = \sqrt{ \frac{2 \times 15}{2.3} } = \sqrt{ \frac{30}{2.3} } \\ = 3.61157559...$

We have the final answer as

Hope this helps you

7 0

4 years ago

El fluoruro de hidrógeno HF que se utiliza en

Blizzard [7]

Answer:

25.6g de HF son producidos

Explanation:

<em>...¿Cuánto HF es producido?</em>

Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:

<em>Moles CaF2:</em>

Masa molar:

1Ca = 40g/mol

2F = 19*2 = 38g/mol

40+38 = 78g/mol

50g CaF2 * (1mol/78g) = 0.641 moles CaF2

<em>Moles H2SO4:</em>

Masa molar:

2H = 2g/mol

1S = 32g/mol

4O = 64g/mol

98g/mol

100g H2SO4 * (1mol / 98g) = 1.02 moles H2SO4

Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.

<em>Moles HF usando la reacción:</em>

0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF

Masa molar:

1g/mol + 19g/mol = 20g/mol

1.282 moles HF * (20g/mol) =

<h3>25.6g de HF son producidos</h3>

8 0

3 years ago

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × $\frac{1mol}{342,15g}$ = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × $\frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH}$ = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × $\frac{2molAl(OH)_{3}}{6 moles NaOH}$ = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

I hope it helps!

3 0

3 years ago