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gtnhenbr [62]
3 years ago
6

Which element in period 3 has three times the electronegativity value compared to that of li in period 2

Chemistry
2 answers:
victus00 [196]3 years ago
7 0

Answer:

Nitrogen

Explanation:

Elements in period two includes lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

According to periodic trends, the electro negativity values are expected to increase across the period up to fluorine. Hence, as we go right wards, we encounter elements with higher electronegative values.

While lithium has an electronegative value of 1 , the electronegative value of element nitrogen is thrrr times this which is equal to three

qaws [65]3 years ago
4 0

Answer:

Chlorine(3.16)

Explanation:

The electronegativity of an element is the ability of an atom to attract a bonding pair of an atom. This measures the tendency of an atom to attract electron to itself in a chemical bond  . One of the most electronegative element is fluorine.  

Electronegativity in the periodic table increases from left to right across the period while it decreases down the group in the periodic table.

Lithium is in period 2 with electronegativity of 0.98 . The element in period 3 with electronegativity times 3 of lithium is chlorine . Chlorine has an electronegativity of  3.16.

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13.2 g of gold

Explanation:

We'll begin by converting 5.25 L to ft³.

This can be obtained as follow:

Recall:

1 L = 0.0353 ft³

Therefore,

5.25 L = 5.25 × 0.0353

5.25 L = 1.85×10¯¹ ft³

From the question given above,

2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.

Therefore, Xg of gold will be present in 1 ft³ of soil i.e

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Ammonium hydrogen sulfide NH4HS(s) decomposes on heating according to the reaction NH4HS(s) ↔ NH3(g) + H2S(g) At 25 °C the equil
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0.66atm

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How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?
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The answer is:  " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂  = 2.0 mol CaCl₂ / L  ; 

Since: "M" = "Molarity" (measurement of concentration); 

                  = moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion:  1000 mL = 1 L . 

Given: 250 mL ;   

250 mL = ?  L  ?  ;  


250 mL * (1 L / 1000 L) =  (250/1000) L = 0.25 L . 
___________________________________________________________
 
(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol  = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ;  Convert to "g" (grams):

→ 0.50 mol CaCl₂  .
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1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl  =  <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;  

→ Note the subscript, "2", in the " Cl₂ " ; 
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So, to calculate the molar mass of "CaCl₂" :

40.08 g  +  2(35.45 g) = 

40.08 g  +  70.90 g = 110.98 g ;  round to 4 significant figures; 

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__________________________________________________________
So:

→  0.50 mol CaCl₂  = ? g CaCl₂  ? ; 

→  0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

                                             = (0.50) * (111 g) CaCl₂ ;

                                             =  55.5 g CaCl₂  ;

                                                → round to 2 significant figures; 

                                                →  56 g CaCl₂ .
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