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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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For the oxidation of ammonia

OverLord2011 [107]

The answer is 0.405 M/s

- (1/3) d[O2]/dt = 1/2 d[N2]/dt

- d[O2]/dt = 3/2 d[N2]/dt

- d[O2]/dt = 3/2 × 0.27

- d[O2]/dt = 0.405 mol L^(-1) s^(-1)

5 0

2 years ago

what would the total pressure of a mixture of fluorine, chlorine, and bromine gases be if the partial pressure are 2.20 atm, 6.7

astraxan [27]

Answer:

The total pressure would be 8, 93 atm

Explanation:

We apply Dalton's laws, where for a gaseous mixture, the total pressure (Pt) is the sum of the partial pressures (Px) of the gases that make up the mixture.

Pt= Pxa + Pxb+ Pxc....

Pt=2, 20 atm+ 6, 70 atm+ 0,03 atm= 8, 93 atm

4 0

3 years ago

Determine whether each molecule is polar or nonpolar.

JulsSmile [24]

A) Sulfur dichloride (SCl₂) is polar molecule because is <span>bent with asymmetric charge distribution around the central atom (S).
b) </span>Sulfur tetrachloride (SCl₄) is polar because there is<span> lone electron pair around the sulfur.
c) </span>Bromine pentachloride (BrCl₅) is polar because dipole moment do not cancel.

6 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

The diagram shows the electron arrangement in a molecule of ammonia, showing only outer

rusak2 [61]

Answer:

Since this is old, im just gonna get these points, don't wan't them to go to waste lm.ao

Explanation:

3 0

3 years ago