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Scrat [10]
3 years ago
6

Consider the elements in Group 14. Which element has the highest electron affinity? A) Carbon B) Germanium C) Lead D) Silicon

Chemistry
1 answer:
svlad2 [7]3 years ago
3 0

Electron affinity is a measure of the tendency of a neutral atom  to gain electrons and form a negative ion. It can be represented by a general equation:

X + e- → X-

The value of electron affinity decreases on moving down a group. This is because on moving down a group the atomic size increases as a result the added electron feels less pull or attraction towards the nucleus.

In group 14, Carbon 'C' is the first member and therefore will have the highest electron affinity.

Ans : A) Carbon

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Mumz [18]
There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.

Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
5 0
2 years ago
Of the five salts listed below, which has the highest concentration of its cation in water? assume that all salt solutions are s
Zigmanuir [339]
First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:

Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively. 

For the solutions ahead, let x be the concentration of the cation.

A.  The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M

B. The formula is Co(OH)₂. 
 1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M

C. The formula is CoS. 
 5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M

D. The formula is Cr(OH)₃. 
 1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M

E. The formula is Ag₂S. 
 6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M

<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>

5 0
2 years ago
What is the difference between accuracy and precision in chemistry ?
earnstyle [38]
Accuracy is more of a range 
6 0
3 years ago
What are the half-reactions for a galvanic cell with Zn and Mg electrodes?
Alona [7]

the half-reactions

cathode : Zn²⁺ (aq) + 2e⁻ ---> Zn (s)  

anode : Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Zn²⁺(aq) + Mg(s)→ Zn(s) + Mg²⁺ (aq)

<h3 /><h3>Further explanation</h3>

Given

Zn and Mg electrodes

Required

The half-reactions for a galvanic cell

Solution

To determine the reaction of a voltaic cell, we must determine the metal that serves as the anode and the metal that serves as the cathode.

To determine this, we can either know from the standard potential value of the cell or use the voltaic series

1. voltaic series

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

So the metal on the left will easily undergo oxidation and function as anode

Since Mg is located to the left of Zn, then Mg functions as anode and Zn as a cathode

2. Standard potentials cell of Mg and Zn metals :

Mg2+ + 2e– → Mg E° = -2,35 V

Zn2+ + 2e– → Zn E° = -0,78 V

The anode has a smaller E°, then Mg is the anode and Zn is the cathode.

7 0
3 years ago
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Elenna [48]

ΔH₁ = ΔH₂ - ΔH₃

ΔH₁ = -635.1-(178.3) KJ

ΔH₁ = -813.4 KJ

3 0
2 years ago
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