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lyudmila [28]
3 years ago
5

Ionic bonds are formed between metal and nonmetal elements. Potassium has only one valence electron and it is stripped away in t

he process of forming an ionic bond. Nonmetals, such as chlorine, tend to gain valence electrons in the formation of ionic bonds. Determine which element(s) are likely to have their electrons stripped away in the formation of ionic bonds. I) aluminum II) sulfur III) oxygen IV) silver V) neon
Chemistry
2 answers:
lisabon 2012 [21]3 years ago
6 0

Answer:

The answer is alimunum and silver.

Explanation:

An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions. Usually, the metal gives up its electrons forming a cation to the nonmetal element, which forms an anion.

As mentioned, metallic elements are those that tend to give up electrons, which bind with atoms that have the ease of accepting them, which are non-metallic elements. Then, <u><em>aluminum and silver, being metals, are the most likely elements to eliminate their electrons in the formation of ionic bonds. </em></u>

Sulfur, oxygen, being nonmetals, will tend to accept electrons. Neon is a noble gas, also called inert, and having its last layer of electrons complete, tends not to react with other elements.

Nonamiya [84]3 years ago
4 0
Aluminum and Silver, as they are metals. Metals are more likely to lose valence electrons.
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Flourine
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2 years ago
Calculate by (a)% weight and (b) %mole each of the elements present in sugar
Musya8 [376]

Explanation:

Molecular mass of sugar = C_{12}H_{22}O_{11} : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Percentage of carbon by weight in C_{12}H_{22}O_{11}:

\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%

Percentage of hydrogen by weight in C_{12}H_{22}O_{11}:

\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%

Percentage of oxygen by weight in C_{12}H_{22}O_{11}:

\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%

b) Percentage of mole each of the elements present in sugar:

=\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in C_{12}H_{22}O_{11}:

\frac{12 mol}{45 mol}\times 100=26.66\%

Percentage of hydrogen by mole in C_{12}H_{22}O_{11}:

\frac{22 mol}{45 mol}\times 100=48.88\%

Percentage of oxygen by mole in C_{12}H_{22}O_{11}:

\frac{11 mol}{45 mol}\times 100=24.44\%

7 0
3 years ago
Magnesium burns with a very bright light. When the flame goes out, a White powder is left behind. what is the word equation for
melisa1 [442]
When magnesium is burned, it reacts with oxygen in air not with the fire. The fire is the energy needed for the reaction to happen. Magnesium reacts with oxygen forming magnesium oxide. The light emitted from the reaction is because the reaction produced a lot of heat.
6 0
3 years ago
Need the names for these asap please ​
jeyben [28]

Answer:

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5 0
3 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands
Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

Energy of 10th orbit = E_{10}

E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J

=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

3 0
2 years ago
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