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prisoha [69]
3 years ago
5

Which explains the change in ionization energy that occurs between removing the first and second electron from an Atom

Chemistry
2 answers:
Mars2501 [29]3 years ago
8 0
 The ionization energy increases because the ratio of the protons to electrons increases. It is quantitatively expressed in symbols as X + energy → X+ + e−.
Tanya [424]3 years ago
6 0

Answer:

Explanation:  Ionization energy is the removal of an electron from the outermost shell of the gaseous atom.

 X(g) \rightarrow  X^{+}(g) + e^{-}

When we remove one electron, the species acquire a positive charge because the number of protons has been increased than the number of electrons . In reference to this, the nucleus would hold its grip to the valence shell more than before due to the electrostatic forces of attraction.

Thus, due to this electrostatic forces of attraction, it will become difficult to remove an electron from the last shell.

Thus ionization energy will increase on removal of successive electrons .

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Notice what happens to the numbers of protons, neutrons, and electrons when you
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Answer:

Only the number of neutrons change.

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A 4kg bowling ball is held 2.5 meters above your head. A 0.5 kg tennis ball is held at 12 meters above your head. Which one has
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Answer:

P.E for the 4kg bowling ball held 2.5 meters above head is 100j, while the other is 60j so the 4kg ball has more potential energy

7 0
3 years ago
KCI is a molecule. True or False
Monica [59]

Answer: duh

Explanation:

4 0
3 years ago
Read 2 more answers
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
How did Rutherford’s atomic model fix the shortcomings of Thomson’s atomic model?
dedylja [7]

<u>Plum Pudding Model(Thomson's atomic model)</u>

  • Thomson's atomic model states that an atom has a positive sphere charge with electrons embedded inside it. He compared the atom with a plum pudding,as the electrons according to him seemed like the dry fruits embedded in the spherical pudding.

<u>Rutherford's Model</u>

  • However Rutherford bombarded high energy streams of α-particles on a thin gold foil of 100 nm thickness. The  deflection produced by  the trajectory of these high energy  α-particles after interaction with the thin sheet of gold was studied by placing a screen made up of zinc sulfide around the gold foil.
  • The major observations made by Rutherford were that  a very huge fraction of α-particles passed through the gold sheet without getting deflected. Thus he concluded that the major part of an atom must be empty.
  • Very few   α-particles  got deflected minutely or at very small angles  by the gold sheet when they were bombarded against it.  Also very few particles got deflected at large angles. This made him conclude that the positive charge is concentrated in a very small region and is  not distributed uniformly.

From the above observations he gave the following postulates:

  • An atom is made up of  positively charged particles. The mass of an atom was concentrated in small region which is  named as the nucleus of an atom.
  • The  nucleus is surrounded by  electrons which are negatively charged particles which  revolve around the nucleus in a fixed circular path called as “orbits.”
  • An atom is  electrically neutral because electrons are negatively charged and the  nucleus is positively charged. The electrons are held by the nucleus due to a  strong electrostatic force.
  • Compared to the total size of an atom the size of the nucleus is very small.
7 0
3 years ago
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