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3 years ago

Which statement is true about the ionic size of the elements in a group as one moves from top to bottom in that group?

Chemistry

2 answers:

nekit [7.7K]3 years ago

8 0

Answer:

Ionic size increases from top to bottom within the group.

Explanation:

Ions are formed when neutral atoms of elements gain or lose electron(s).
The ionic radius is the distance from the nucleus of an ion to the outermost energy level.
The ionic size or radius increases down the group as the number of energy level increases.
Therefore, an ion of an element lower in the group will be larger than the ion of an element higher in the group.
For example, an ion of potassium, K+, is larger in size compared to the ion of sodium, Na+ since K+ has more energy levels than Na+.

Maru [420]3 years ago

8 0

Answer:

Option B = ionic sizes increases from top to bottom.

Explanation:

Definition of atomic radii:

The atomic radius is the distance between center of two bonded atoms.

Cation:

When an atom lose the electrons the formed is called cation.

e.g Na⁺ sodium lose one electron and form cation.

Anion:

when an atom gain the electrons the ions formed is called anion.

e.g N³⁻

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase.The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases. Same is the case with ions. Ionic size decreases moving from left to right.

Trend along group:

In group by addition of electron atomic radii increase from top to bottom due to increase in atomic number and addition of extra shell. The ionic size also increases from top to bottom.

If we compare the cationic and anionic size with neutral atomic radii, the anionic size is greater than neutral atomic radii because of addition of extra shell while cationic size is smaller because of removal of one electronic shell.

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A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon

svetlana [45]

Answer:

$\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}$

Explanation:

1. Miles travelled in an average month

$\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}$

2. Using a gasoline powered vehicle

(a) Moles of heptane used

$n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}$

(b) Equation for combustion

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O

(c) Moles of CO₂ formed

$n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}$

(d) Volume of CO₂ formed

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.

$V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}$

3. Using an electric vehicle

(a) Theoretical energy used

$\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}$

(b) Actual energy used

The power station is only 85 % efficient.

$\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\$

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

The heat of combustion of methane is -802.3 kJ·mol⁻¹

$V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}$

4. Comparison

$\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}$

6 0

3 years ago

What is the average binding energy per nucleon for a U-238 nucleus with a mass defect of 0.184 amu? (1 amu= 1.66 x 10-27 kg; 1 J

Bond [772]

add all the number and find the average then subtract the mass defect and then you will get your answer

3 0

3 years ago

ANSWER FAST PLEASE!<br> Balance the following equation:<br> __Hg + __O₂ --> __HgO

ICE Princess25 [194]

Answer:

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four Mercury + four oxygen

3 0

3 years ago

Propane (c3h8) is burned in oxygen to produce carbon dioxide and water. the heat of combustion of propane is -2012 kj/mole. how

Olegator [25]

$C_{3} H_{8} + 5 O_{2} ---\ \textgreater \ 3CO_{2} +4H_{2}O (-2012 \frac{kJ}{mol} )


3 mol 10 mol


C_{3}H_{8} is /excess /reactant

because 3 mol propane require 15 mol oxygen (by reaction)

5 mol oxygen ---1 mol propane, so

10 mol oxygen ---2 mol propane

Only 2 mole propane will be burned,

so 

2012 ( kJ/mol)*2 mol =4024 KJ heat will be given off

Correct answer is number 4.


$

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3 years ago

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topjm [15]

Answer:

0.0000098 should be the answer

Explanation:

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