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GREYUIT [131]
3 years ago
6

Determine the differences In electronegativity ionic radius atomic radius and first ionization energy for oxygen and beryllium

Chemistry
1 answer:
Vilka [71]3 years ago
7 0

Answer:

Electronegativity  = 1.87.

Ionic radius = 109 pm.

Atomic radius = -39 pm

First ionization energy = 410 kJ/mol

Explanation:

Hello!

In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.

In such a way, the required comparison is written below:

Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.

Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.

Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm

First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol

It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.

Best regards!

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4. A galvanic cell is formed when two metals are immersed in solu- tions differing in concentration 1 when two different metals
emmasim [6.3K]

A galvanic cell is formed when two metals are immersed in solutions differing in concentration 1 when two different metals are immersed.

<h3>What is galvanic cell?</h3>

A galvanic cell is an electrochemical device that transforms chemically generated free energy into electrical energy. A photogalvanic cell produces photochemical species that react to produce an electrical current when connected to an external circuit.

<h3>How does galvanic cell works?</h3>

In order to create a pathway for the flow of electrons via this wire, the galvanic cell makes use of the ability to split the flow of electrons during the processes of oxidation and reduction, forcing a half-reaction and linking each with a wire.

An electrochemical device known as a galvanic cell converts chemical energy from a spontaneous redox response into electrical energy. It possesses an electrical potential of 1.1 V. The anode, which is a negative plate in galvanic cells, is where oxidation takes place. It is a positive plate where lessening takes place.

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8 0
2 years ago
Charlotte is driving at 58.6 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 4
kramer

Answer:   19.71 feet

Explanation:

Given:  Speed of Charlotte =58.6 mi/h

Since 1 hour = 3600 seconds

and 1 mile = 5280 feet

So, Speed of Charlotte = \dfrac{58.6\times5280}{3600}\approx85.95 mi/ sec

She looks down at her phone and takes her eyes off the road for 4.36 s.

Since , Distance = \dfrac{speed}{time}

So, Distance = \dfrac{85.95}{4.36}=19.71\ feet

Hence, Charlotte traveled  19.71 feet during this time.

8 0
3 years ago
Which example is the site of reduction when the dry cell is operating
Simora [160]

Answer:

A battery contains electrochemical cells that can store chemical energy to be converted to electrical energy. A dry-cell battery stores energy in an immobilized electrolyte paste, which minimizes the need for water. Common examples of dry-cell batteries include zinc-carbon batteries and alkaline batteries.

Explanation:  i hope this helps sorry if it didnt

4 0
3 years ago
Read 2 more answers
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

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7 0
2 years ago
How many moles of water are there in 1.000 l at stp? assume the density of water is 1000 kg/m3?
EastWind [94]
Assume 1 liter = 1 kilogram of water = 1000 grams of water.

Part A)
MW of hydrogen is 1.008g/mol, and oxygen is 16.00g/mol.

Find the MW of water by
2*(1.008) + (16.00) = 18.016g/mol.

Convert 1000g H2O to moles :

(1000g H2O)*(1mol H2O / 18.016g H2O) = 55.51 mol

Part B)
Using the answer from part A and Avogadro's number:

(55.51mol)*(6.022*10^23) =

3.343*10^25 molecules.
Hope this is helpful
7 0
3 years ago
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