What is the change of a liquid to a solid, 15 letter word, 4th letter is "s".

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

egoroff_w [7]

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.

5 0

3 years ago

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Can we cover the earth with one mole of pennies?

lara [203]

Answer:

\no

Explanation:

6 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.