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elixir [45]
2 years ago
14

As the elements in Period 2 of the Periodic Table are considered in succession from left to right, there is a decrease in atomic

radius with increasing atomic number. This may best be explained by the fact that the
a. number of protons increases, and the number of shells of electrons remains the same
b. number of protons increases, and the number of shells of electrons increases
c. number of protons decreases, and the number of shells of electrons remains the same
d. number of protons decreases, and the number of shells of electrons increases
Chemistry
2 answers:
maria [59]2 years ago
7 0

Answer:

A. number of protons increases, and the number of shells of electrons

remains the same.

Explanation:

The number of protons increases as you go from left to right in a period. That increases the number of positive charges in the nucleus. The number of negatively charged electrons in the last shell also increases but the number of shells remains the same. The increasingly positive nuclear charge attracts the increasingly negative shells. The result is the atomic radius decreases as the atomic number increases in a period.

Art [367]2 years ago
4 0
B)the number of protons increases ,and the number of shells of electrons increases
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Recalling that a beaker of water is two dimensional, what is the three dimensional shape of the micelle
saul85 [17]
A micelle refers to an aggregate of a surfactant molecule, which a dispersed in a liquied colloid. In an aqueous liquid, a micelle is arranged in such a way that its hydrophillic head region will be in contact with the surrounding solvents while the hydrophobic tail region will be embedded in its center. 
The three dimensional shape of a micelle is A SPHERE. 
7 0
2 years ago
Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give
denpristay [2]
Entropy Change is calculated  by (Energy transferred) / (Temperature in kelvin) 
deltaS = Q / T 

Q = (mass)(latent heat of fusion) 
Q = m(hfusion) 
Q = (500g)(333J/g) = 166,500J 

T(K) = 32 + 273.15 = 305.15K 
deltaS = 166,500J / 305.15K 
deltaS = 545.63 J/K
3 0
2 years ago
calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c
Pepsi [2]

<u>Answer:</u>

<em>The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)</em>

<u>Explanation:</u>

We make use of the formula

Q=m \times c \times \Delta T

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

Q=m \times c \times \Delta T

Rearranging the formula

c= \frac {Q}{(m\times \Delta T)}

= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}

So,

c = 0.129 J/(g℃)

(Answer)

7 0
2 years ago
When iron pyrite (FeS2) is heated in air, the process known as "roasting" forms sulfur dioxide and iron(III) oxide. When the equ
Ronch [10]

Answer:

The coefficient of O2 is 11

Explanation:

Step 1:

The equation for the reaction:

FeS2 + O2 → SO2 + Fe2O3

Step 2:

Balancing the equation. The equation can be balance as follow:

FeS2 + O2 → SO2 + Fe2O3

There are 2 atoms of Fe on the right side and 1 atom on the left. It can be balance by putting 2 in front of FeS2 as shown below:

2FeS2 + O2 → SO2 + Fe2O3

There are 4 atoms of S on the left side and 1 atom on the right side. It can be balance by putting 4 in front of SO2 as shown below:

2FeS2 + O2 → 4SO2 + Fe2O3

Now, there are a total of 11 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 11/2 in front of O2 as shown below:

2FeS2 + 11/2O2 → 4SO2 + Fe2O3

Multiply through by 2 to clear the fraction as shown below:

4FeS2 + 11O2 → 8SO2 + 2Fe2O3

Now the equation is balanced.

The coefficient of O2 is 11

8 0
2 years ago
Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O
oee [108]

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

<u>To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺</u>:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the <u>number of Cr atoms</u> on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, <u>6 electrons are gained</u> by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to <u>balance the total charge</u> on the reactant and product side,<u> 14 H⁺ is added on the reactant side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to <u>balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

<u>Therefore, the correct balanced reduction half-reaction is:</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

3 0
2 years ago
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