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Goryan [66]
3 years ago
5

Atoms of metallic elements can form ionic bonds, but they aren't very good at forming covalent bonds. why?

Chemistry
1 answer:
Solnce55 [7]3 years ago
7 0
<span>They have a great tendency to lose electrons</span>
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The competition between the U.S. and the Soviet Union to get satellites and humans into space was known as the:
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Answer: It was known as the SPACE RACE.

Explanation: The space race was a cold war between the Soviet Union (USSR) and the United States (US), to get satellites and humans into space. The competition began in earnest on August 2, 1955, when the Soviet Union responded to the US announcement. On April 12, 1961, the USSR surprised the world again by launching Yuri Gagarin into a single orbit around the Earth in a craft they called Vostok 1.

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What is the mass of 3.75 moles of NaCl
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Mass of salt is equal to 218.8125 grams.

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All of the following are benefits of recycling except?
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Recycling Isn’t Always Cost Effective.
High Up Front Cost
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3 years ago
Draw the conjugate base for the Brønsted-Lowry acid-base reaction that occurs when the following acid reacts with water. Show al
DanielleElmas [232]

Answer:

The structures are shown below.

Explanation:

When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.

The formal charge of an atom can be calculated by:

FC = X - (Y + Z/2)

Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.

a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.

b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.

c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.

The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.

The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.

The structure is shown in figure c.

7 0
3 years ago
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Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)
Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0
3 years ago
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