Physics, 05.12.2020 20:20 Bowied6025

I don't know if this is the right category or not but I don't care. Ozone consists of three oxygen atoms (O3) and water consists of two hydrogen atom and one oxygen atom (H2O).

Compare water (H2O) to a single atom of oxygen (O). Is each an element or a compound? Explain.

I don't know if this is the right category or not but I don't care.

Ozone consists of three oxyge

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Answer from: LigmaXD

Forming a covalent bond

A covalent bond is formed when two atoms share a pair of electrons. Covalent bonding occurs in most non-metal elements, and in compounds formed between non-metals.

These shared electrons are found in the outer shells of the atoms. Usually each atom contributes one electron to the shared pair of electrons.

The slideshow shows how a covalent bond forms between a hydrogen atom and a chlorine atom, making hydrogen chloride.

Structures of a hydrogen atom and a chlorine atom.

1. A hydrogen atom with one electron and a chlorine atom with 17 electrons

Molecules

Most covalently bonded substances consist of small molecules. A molecule is a group of two or more atoms joined together by covalent bonds. Molecules of the same element or compound always contain the same number of atoms of each element.

The atoms in a molecule are always joined together by a covalent bond. Substances that are made up of ions do not form molecules.

Sizes of atoms and simple molecules

A small molecule contains only a few atoms, so atoms and small molecules have a similar range of sizes. They are very small, typically around 0.1 nm or 1 × 10-10 m across.

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Answer from: Quest

ice to water 1 water to vapour 2 latent heat

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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