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topjm [15]
3 years ago
12

When atoms from Column I (Group 1) combine with atoms from Column VII (Group 17): A nearly 100% covalent bond forms. The bond wi

ll be ionic. A weak bond will form. The resulting compound will be very unstable. A great deal of energy is absorbed.
Chemistry
2 answers:
gtnhenbr [62]3 years ago
7 0

The answer you're looking for is B. The bond will be ionic.

kotykmax [81]3 years ago
6 0

Answer:

When atoms from Column I (Group 1) combine with atoms from Column VII (Group 17) <u><em>the bond will be ionic.</em></u>

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.

In other words, by forming ionic compounds, the elements to the left of the periodic table usually lose electrons, and form a cation that has the same electronic configuration as the nearest noble gas. The elements to the right of the periodic table tend to gain electrons to reach the electronic configuration of the next noble gas .

This type of bond is produced by the attraction of ions with opposite charges. By assigning or accepting the valence electrons, the ions comply with the octet rule and are more stable. Remember that valence electrons are the electrons of the outermost layer of the atom, which can participate in chemical bonds.

It is also possible to think that an ionic bond is formed when two atoms have a very large electronegativity difference, remembering that electronegativity is the property of an atom to attract an electron.  And this large difference in electronegativity usually occurs between the atoms of Column I (Group 1) combined with atoms of Column VII (Group 17).

Then it is finally possible to say that <u><em>"When atoms from Column I (Group 1) combine with atoms from Column VII (Group 17) the bond will be ionic."</em></u>

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Answer:

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Explanation:

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Do you no the 5 major dairy groups​
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Question 1(Multiple Choice Worth 4 points)
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<u>Answer </u>

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of H_{2}O over Nine moles of O_{2}

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of Fe_{2}O_{3}.

Answer 4 : Mass of O_{2} = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

<u>Solution </u>

Solution 1 : Given,

Given mass of Fe_{2}O_{3} = 55 g

Molar mass of Fe_{2}O_{3} = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of Fe_{2}O_{3} = \frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}} = \frac{55 g}{159.69 g/mole} = 0.344 moles

Balanced chemical reaction is,

Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)

From the given reaction, we conclude that

1 mole of Fe_{2}O_{3} gives              →         3 moles of CO

0.344 moles of Fe_{2}O_{3} gives    →         3 × 0.344 moles of CO

                                                     =         1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

                    = 1.032 × 28.01

                    = 28.90 g

Solution 2 : The balanced chemical reaction is,

2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O

From the given reaction, we conclude that the Six moles of H_{2}O over Nine moles of O_{2} is the correct option.

Solution 3 : The balanced chemical reaction is,

4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of Fe_{2}O_{3}.

Solution 4 : Given,

Given mass of Zn(ClO_{3})_{2} = 150 g

Molar mass of Zn(ClO_{3})_{2} = 232.29 g/mole

Molar mass of O_{2} = 31.998 g/mole

Moles of Zn(ClO_{3})_{2} = \frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}} = (\frac{150\times 1}{232.29})moles

The balanced chemical equation is,

Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}

From the given balanced equation, we conclude that

1 mole of Zn(ClO_{3})_{2} gives          →       3 moles of O_{2}

(\frac{150\times 1}{232.29})moles of Zn(ClO_{3})_{2} gives  →  [(\frac{150\times 1}{232.29})\times 3] moles of O_{2}

Mass of O_{2} = Number of moles of O_{2} × Molar mass of  O_{2} = [(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams

Therefore, the mass of O_{2} = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of Na_{2}SO_{4} = 4.2 moles

Balanced chemical equation is,

H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}

From the given chemical reaction, we conclude that

1 mole of Na_{2}SO_{4} obtained from 2 moles of NaCN

4.2 moles of Na_{2}SO_{4} obtained   →   2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles


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