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Over [174]
3 years ago
14

Based upon the following diagram, propose a possible identity for atoms X and Y. Explain your answer in terms of the periodic ta

ble and ionic bonding. Lastly, explain why the atoms bond in this ratio.

Chemistry
1 answer:
zhenek [66]3 years ago
8 0

Answer:

Up until now we have been discussing only the elemental forms of atoms which are neutrally charged. This is because the number of electrons (negative in charge) is equal to the number of protons (positive in charge). The overall charge on the atom is zero, because the magnitude of the negative charge is the same as the magnitude of the positive charge. This one-to-one ratio of charges is not, however, the most common state for many elements. Deviations from this ratio result in charged particles called ions.

Throughout nature, things that are high in energy tend to move toward lower energy states. Lower energy configurations are more stable, so things are naturally drawn toward them. For atoms, these lower energy states are represented by the noble gas elements. These elements have electron configurations characterized by full s and p subshells. This makes them stable and unreactive. They are already at a low energy state, so they tend to stay as they are.

The elements in the other groups have subshells that are not full, so they are unstable when compared to the noble gases. This instability drives them toward the lower energy states represented by the noble gases that are nearby in the periodic table. In these lower energy states, the outermost energy level has eight electrons (an “octet”). The tendency of an atom toward a configuration in which it possesses eight valence electrons is referred to as the “Octet Rule.”

There are two ways for an atom that does not have an octet of valence electrons to obtain an octet in its outer shell. One way is the transfer of electrons between two atoms until both atoms have octets. Because some atoms will lose electrons and some atoms will gain electrons, there is no overall change in the number of electrons, but with the transfer of electrons the individual atoms acquire a nonzero electric charge. Those that lose electrons become positively charged, and those that gain electrons become negatively charged. Recall that atoms carrying positive or negative charges are called ions. If an atom has gained one or more electrons, it is negatively charged and is called an anion. If an atom has lost one or more electrons, it is positively charged and is called a cation. Because opposite charges attract (while like charges repel), these oppositely charged ions attract each other, forming ionic bonds. The resulting compounds are called ionic compounds.

The second way for an atom to obtain an octet of electrons is by sharing electrons with another atom. These shared electrons simultaneously occupy the outermost shell of both atoms. The bond made by electron sharing is called a covalent bond. Covalent bonding and covalent compounds will be discussed in Chapter 4 “Covalent Bonding and Simple Molecular Compounds”.

At the end of chapter 2, we learned how to draw the electron dot symbols to represent the valence electrons for each of the elemental families.  This skill will be instrumental in learning about ions and ionic bonding. Looking at Figure 3.1, observe the Noble Gas family of elements. The electron dot symbol for the Nobel Gas family clearly indicates that the valence electron shell is completely full with an octet of electrons.  If you look at the other families, you can see how many electrons they will need to gain or lose to reach the octet state.  Above, we noted that elements are the most stable when they can reach the octet state. However, it should also be noted that housing excessively high negative or positive charge is unfavorable.  Thus, elements will reach the octet state and also maintain the lowest charge possible.   You will note that for the IA, IIA, IIIA and transition metals groups, it is more economical to lose electrons (1-3 electrons) from their valence shells to reach the octet state, rather than to gain 5-7 electrons.  Similarly main group columns VA, VIA, and VIIA tend to gain electrons (1-3) to complete their octet, rather than losing 5-7 electrons. Some atoms, like carbon, are directly in the middle.  These atoms don’t like to gain or lose electrons, but tend to favor the sharing model of chemical bonding. The remaining sections of this chapter will focus on the formation of ions and the resulting ionic compounds.

Explanation:

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Convert 4.92 L of hydrogen at STP to moles.
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{\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}

\sf C) \:  0.220 \: mol \: H_2

Explanation:

1 mol = 22.4 litres

4.92 = how many moles?

\sf  \implies  \frac{4.92}{22.4}  \\  \\  \sf  \implies 0.2196 \\  \\  \sf  \implies 0.220 \: mol \: H_2

5 0
2 years ago
Read 2 more answers
What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed
monitta

Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

                                     = 32.042 g/mol

Molar Mass of CO₂      = 1(12.01 g/mol) + 2(16.00 g/mol)  

                                     = 44.01 g/mol

                                   

Mass of only one reactant i.e. CH₃OH is given so  it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

                                     10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

                          1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

                             2 mol CO₂ / 2 mol CH₃OH

Convert moles of  CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

                           44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

                       = 0.707 x 100%

                       = 70.7 %

Hence option A 70.7% yield is the correct answer.

8 0
4 years ago
After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage
Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have <u>2 reactions</u>, the formation of the <u>grignard reagent</u> and the <u>formation of the alcohol</u>. The first step then is the calculation of the <u>maximum amount</u> of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the <u>following conversion ratios</u>:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide (C_6H_5Br)= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and  C_6H_5Br= 1:1

2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg

9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br

The smallest value is the mol of bromobenzene therefore <u>0.0402 mol</u> of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of <u>formation of the alcohol</u>. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

4.83~g~\frac{1~mol}{136.14~g}=0.35~mol

The we have to <u>divide by the coefficient</u> of each reactive in the balance reaction. So:

\frac{0.35~mol}{1}=0.35

\frac{0.0402~mol}{2}=0.0201

Therefore the <u>limiting reagent</u> would be the phenylmagnesium bromide. Now, the <u>molar ratio</u> between the phenylmagnesium bromide and triphenyl carbinol is <u>2:1</u>, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to <u>grams of triphenyl carbinol</u>:

Molar mass of triphenyl carbinol= 260.33 g/mol

0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol

Finally, we have to <u>divide</u> the obtanied solid by the calculated one:

Percentage=\frac{5}{5.23}*100=95.6\%

5 0
3 years ago
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