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balandron [24]
3 years ago
9

Which is the correct name of a compound that is formed by Si and O? silicon dioxide silicon trioxygen disilicon dioxygen monosil

icon oxide
Chemistry
2 answers:
Flauer [41]3 years ago
8 0

The correct is tricky, be careful. The right is silicon dioxyde (SiO2)

Silicon Oxides are written in the form SiOx, (0 <x <2), so:

there is no silicon trioxygen and disilicon dioxygen.

SiO is called silicon monoxide and not monosiicon oxygen, so this proposition is false.

All that remains is the silicon dioxide (SiO2) that is written correctly.

Silicon dioxide can be synthesized but also exists in abundance in nature. Silicon (Si) represents about 26% of the Earth's crust. Silica (SiO2), the natural form of silicon dioxide, accounts for about 60%.

Natalka [10]3 years ago
7 0

Answer:silicon dioxide

Explanation:

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The atomic number,atomic mass,valence electrons and the formula for calculating the number of neutrons is also given.The following table has been completed below in the attachment.

Explanation:

Atomic Number:the number of protons /electrons present in the nucleus of an atom is called the atomic number.

Mass Number/Atomic Mass:the number of protons and neutrons present in a nucleus

<u>Example:</u>C_{6} ^{12};here;6 implies the atomic number and 12 implies the mass number/atomic mass

<u><em>No.of neutrons=Mass number-Number of protons</em></u>

Valence electrons:the number of electrons present in the outer most shell of an atom are called valence electrons

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A sample of Cd(OH)2 is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of Cd +2 = 1.7 x 10 -5M
Archy [21]

Answer:

Ksp=2.0x10^{-14}

Explanation:

Hello there!

In this case, given the solubilization of cadmium (II) hydroxide:

Cd(OH)_2(s)\rightleftharpoons Cd^{2+}(aq)+2OH^-(aq)

The solubility product can be set up as follows:

Ksp=[Cd^{2+}][OH^-]^2

Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:

Ksp=(1.7x10^{-5})(3.4x10^{-5})^2\\\\Ksp=2.0x10^{-14}

Regards!

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