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dusya [7]
2 years ago
7

The properties of two elements are listed below. Element Atomic radius (pm) Ionic radius (pm) First ionization energy (kJ/mol) E

lectron affinity (kJ/mol) Electronegativity Br 114 195 1140 –325 3.0 K 243 152.2 418.8 –48.4 0.82 Which prediction is supported by the information in the table? K will give up an electron more easily than Br. Both K and Br will have the same pull on electrons. K will have a smaller size in comparison to Br. Both K and Br will produce ions of the same size.
Chemistry
2 answers:
kifflom [539]2 years ago
5 0

Based on the data presented in the table and comparing it to the given statements we can interpret that potassium (k) will give up electrons readily than bromine (Br)

Now, ionization energy is the amount of energy required to remove an electron from an atom or ion. In the case of K the first IE = 418.8 kJ/mol while for Br, IE = 1140 kj/mol. Thus the ionization energy of Br >> K which supports the prediction that K will give up an electron more easily than Br.

Helen [10]2 years ago
3 0
The First option may be correct. Im deeply sorry if im wrong!

FIRST OPTION (A)(1)
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A 0.3870-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7191Â g of co
olchik [2.2K]
The chemical formula for the compound can be written as,

    CxHyOz 

where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
   
    CxHyOz + O2 --> CO2 + H2O 

number of moles of C:
     (0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
 This signifies that 0.0163 mole of C and the mass of carbon in the compound,
        (0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C

number of moles H:
      (0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O

This signifies that there are 0.01635 atoms of H in the compound.
      mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H

Mass of oxygen in the compound,
   0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g

Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O

The formula of the compound is,
      C0.0163H0.01635O0.0109

Dividing the numbers by the least number,
    C3/2H3/2O

The empirical formula of the compound is therefore,
    <em>  C₃H₃O₂</em>
5 0
1 year ago
Draw Lewis structures for each of the following moecules. Show resonance structures, if they exist
Svetllana [295]

 A Lewis structure is a visual representation of the bonds between atoms and it shows the lone pairs of electrons in molecules. This structure is also referred as Lewis dot diagram. Resonance structure are multiple Lewis structure that describe a single molecule.<span>
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6 0
1 year ago
Mrs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she
Mars2501 [29]

Answer:

H2SO, and CaCO

Explanation:

8 0
2 years ago
Read 2 more answers
At which of the following temperatures is the average kinetic energy of the molecules of a substance the least? 23.2 °C 35.2 °C
svp [43]
At 15.2°C. Kinetic energy of molecules highly depends on the temperature — the warmer it is, the faster the molecules will move, especially in fluids (gases and liquids). If we consider that the formula for average kinetic energy of molecules is:

Ek = 3/2*k*T where k is Boltzmanns constant and 3/2 is, well, 3/2, kinetic energy of molecules really only depends on the temperature.
7 0
2 years ago
How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:
WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

                   CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

                       CS₂   +    3Cl₂  -------->   CCl₄   +  S₂Cl₂

                       1 mol     3 mol               1 mol      1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of  S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 +  2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

             CS₂              +    3Cl₂             -------->   CCl₄   +  S₂Cl₂

      1 mol (76 g/mol)     3 mol (71 g/mol)

                CS₂   +    3Cl₂   -------->   CCl₄   +  S₂Cl₂

                 76 g       213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

                   CS₂    :    Cl₂

                   76 g   :    213 g

                    1  g     :      2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

                 2.8 g of Cl₂ ≅  1 g of CS₂

                 10 g of Cl₂ ≅  ? g of CS₂

by doing cross multiplication

                    g of CS₂ = 10 x 1 / 2.8

                    g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

               Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

 

8 0
1 year ago
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