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Elden [556K]
3 years ago
5

1. Select the more electronegative element in this pair. a. F b. Br 2. Select the more electronegative element in this pair. a.

Se b. O 3. Select the more electronegative element in this pair. a. Si b. N 4. Select the more electronegative element in this pair. a. Mg b. Na
Chemistry
2 answers:
dimulka [17.4K]3 years ago
7 0

Answer: 1. F 2. O 3. Mg

Explanation: For electronegativity the trend are as follows.

Moving up group increases the electronegativity so F is in Group 7 together with Br. O and Se is in group 6. For Na and Mg the trend for elements in the same periods is electronegativity increases from left to right.

nignag [31]3 years ago
5 0

Answer:

F

O

N

Mg

Explanation

The key explanation here is that we need to know the trends in electronegativity value increase or decrease across the periods and down the group.

Across the periods, electronegativity is expected to increase. Hence, with the exception of the noble gases, the most electronegative elements are found further right on the periodic table. While electronegativity is expected to increase across the period, it decreases down the group.

Hence, if two elements are in the same group, further down we have the one with the less electronegative value. If two elements are in the same period, further right we have the more electronegative element.

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20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as
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The reaction forms 98.76 g AlCl_3.  

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.  

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

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Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of AlCl_3 we can obtain from each reactant.  

<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

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3 years ago
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