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3 years ago

The tendency of an element to react is closely related to

1 answer:

4 0

Answer: The tendency of an element to react is closely related to the number of valence electrons in the element.

Explanation:

Chemical reactivity is defined as the tendency of an element to loose of gain electrons.

Metals tend to loose electrons and so their chemical reactivity is the tendency to loose electrons.

Non-metals tend to gain electrons and so their chemical reactivity is the tendency to gain electrons.

The number of electrons that an element will loose or gain depends on the number of valence electrons present around that element.

For Example: Chlorine has 7 valence electrons and need 1 electron to complete its octet, whereas sulfur has 6 valence electrons and need 2 electrons to complete its octet.

So, chlorine will gain 1 electron easily than sulfur and thus, is more reactive than sulfur.

Hence, the tendency of an element to react is closely related to the number of valence electrons in the element.

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How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?

Tom [10]

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 .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl = 35.45 g .

There are 2 atoms of Cl in " CaCl₂ " ;

→ Note the subscript, "2", in the " Cl₂ " ;
__________________________________________________________
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;

→ round to 111 g/mol .
__________________________________________________________
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 .
___________________________________________________________
The answer is: " 56 g CaCl₂ " .
___________________________________________________________

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Which of the following is NOT true about equilibrium?

podryga [215]

Answer:

Explanation:

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Is there any way the pandemic helps this society? <img src="https://tex.z-dn.net/?f=22%20%5Ctimes%2012" id="TexFormula1" titl

Alexxandr [17]

Answer:

u meant the answer to this ques

Explanation:

×12

------

22 ×

----------

264

-----------

or the answer can be 22×12=264

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Which organisms are the best example of producers in a food web? A. zooplankton B. krill C. killer whales D. marine phytoplankto

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D. Marine phytoplankton

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How is burning magnesium different than burning methane

olya-2409 [2.1K]

You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.

Explanation:

To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:

2 Mg (s) + O₂ (g) → 2 MgO + heat

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat

However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).

Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)

2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)

So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.

we have used the following notations:

(s) - solid

(g) - gas

(l) - liquid

Learn more about:

combustion reactions

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3 years ago