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m_a_m_a [10]
3 years ago
7

Select the keyword or phrase that will best complete each sentence about the three general rules of for drawing Lewis structures

.a. main group b. hafnium c. valence d. hydrogen e. core f. transition metal1. Draw only__the electrons. 2. Give every___element (except hydrogen) an octet of electrons. 3. Give each____two electrons.
Chemistry
1 answer:
notsponge [240]3 years ago
4 0

Answer:

1. Draw only the valence electrons.

2. Give every element main group element (except hydrogen) an octet of electrons.

3. Give each hydrogen two electrons.

Explanation:

Lewis structures are used to describe the arrangement or configurations of the valence electrons of molecules and polyatomic ions involved in electronic bonding. A Lewis structure consists of the symbols of the elements in the molecule surrounded by dots with each dot representing each of the elements valence electrons. Also, the electrons shared between two elements are shown by dots between the two elements and these electrons are known as shared electron pairs. The valence electrons on atom that is not involved in bonding is known as lone pairs.

The three general rules for drawing Lewis structures are:

1. Draw only the valence electrons. Only the valence electrons of the atoms of elements are shown since they are the only electrons involved in chemical bonding.

2. Give every element main group element (except hydrogen) an octet of electrons. The complete eight valence electrons of the noble gases is associated with their stability. Thus, the main group elements show a tendency to form enough bonds to obtain eight valence electronsmin order to achieve stability. This is known as the octet rule. However, since the maximum number of valence electrons for elements in the first period of the period table is two, the noble gas helium has completely-filled valence shell containing two electrons known as a duplet. Hydrogen belongs to the first period and is therefore an exception tomthe octet rule.

3. Give each hydrogen two electrons. Hydrogen attains a duplet structure in accordance with the structure of helium

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How many grams are in 1 mole of Pb3(PO4)4?
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In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams

Explanation:

This compound is the lead (IV) phosphate.

Grams that occupy 1 mole, means the molar mass of the compound

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621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m

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2 years ago
1.86 g H2 is allowed to react with 9.75 g N2 , producing 2.87g NH3.
svet-max [94.6K]

Answer:

                     (a)  Theoretical Yield  =  10.50 g

                      (b)   %age yield  = 27.33 %

Explanation:

Answer-Part-(a)

                 The balance chemical equation for the synthesis of Ammonia is as follow;

                                          N₂ + 3 H₂ → 2 NH₃

Step 1: Calculating moles of N₂ as;

                   Moles = Mass / M/Mass

                   Moles = 9.75 g / 28.01 g/mol

                   Moles = 0.348 moles of N₂

Step 2: Calculating moles of H₂ as;

                   Moles = Mass / M/Mass

                   Moles = 1.86 g / 2.01 g/mol

                   Moles = 0.925 moles

Step 3: Finding Limiting reagent as;

According to equation,

                1 mole of N₂ reacts with  =  3 moles of H₂

So,

             0.348 moles of N₂ will react with  =  X moles of H₂

Solving for X,

                     X = 3 mol × 0.348 mol / 1 mol

                     X = 1.044 mol of H₂

It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen  is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculating moles of Ammonia as,

According to equation,

                3 mole of H₂ produces  =  2 moles of NH₃

So,

             0.925 moles of H₂ will produce  =  X moles of NH₃

Solving for X,

                     X = 2 mol × 0.925 mol / 3 mol

                     X = 0.616 mol of NH₃

Step 5: Calculating theoretical yield of Ammonia as,

                     Theoretical Yield  =  Moles × M.Mass

                     Theoretical Yield  =  0.616 mol  × 17.03 g/mol

                     Theoretical Yield  =  10.50 g

Answer-Part-(b)

                    %age yield  = Actual Yield / Theoretical Yield × 100

                    %age yield  = 2.87 g / 10.50 g × 100

                    %age yield  = 27.33 %

4 0
3 years ago
Determine how many gmol, kmol, and lbmols there are in 50 kilograms of n-hexane.
Artist 52 [7]

Answer: 581 gmol

0.581 kmol

1.28\times 10^{-3}lbmol

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50\times 1000g}{86g/mol}=581mol

1. The conversion for mol to gmol

1 mol = 1 gmol

581 mol= \frac{1}{1}\times 581=581gmol

2. The conversion for mol to kmol

1 mol = 0.001 kmol

581 mol= \frac{0.001}{1}\times 581=0.581kmol

3. The conversion for mol to lbmol

1 mol = 2.2\times 10^{-3}lbmol

581 mol= \frac{2.2\times 10^{-3}}{1}\times 581=1.28\times 10^{-3}lbmol

3 0
3 years ago
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