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podryga [215]
3 years ago
5

Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium

atom. Express your answers as chemical equations separated by commas. Identify all of the phases in your answer.
Chemistry
1 answer:
Inessa [10]3 years ago
5 0

Answer:

Gd(g) → Gd^{+}(g) + e^{-} , Gd^{+}(g)→ Gd^{2+}(g) + e^{-}, Gd^{2+}(g) → Gd^{3+}(g) + e^{-}

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ Gd^{+}(g) + e^{-}

The second ionization energy is the energy required to remove next/second electron from x^{+} ion. The second ionization energy is always higher than the first:

Gd^{+}(g) → Gd^{2+}(g) + e^{-}

The third ionization energy is the energy required to remove third electron from x^{2+} ion:

Gd^{2+}(g) → Gd^{3+}(g) + e^{-}  

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An ice cute at 0.0*C was dropped into 30.0 g of water in a
PilotLPTM [1.2K]

<u>Answer:</u> The mass of ice cube is 77.90 grams

<u>Explanation:</u>

When ice is mixed with water, the amount of heat released by ice will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

m_1 = mass of ice = ? g

m_2 = mass of water = 30.0 g

T_{final} = final temperature = 19.5°C

T_1 = initial temperature of ice = 0.0°C

T_2 = initial temperature of water = 45.0°C

c_1 = specific heat of ice = 2.108  J/g°C

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

m_1\times 2.108\times (19.5-0)=-[30.0\times 4.186\times (19.5-45.0)]

m_1=77.90g

Hence, the mass of ice cube is 77.90 grams

5 0
3 years ago
What is the new temperature when 10 L at 5 K is compressed to 4.00 L
lina2011 [118]

Answer: C

Explanation:

For this problem, we would use Charles's Law. We fill in what we are given and solve.

Charles's Law: \frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }

\frac{10 L}{5K} =\frac{4.00L}{T_{2} }

10T_{2}= 20

T_{2}=2K

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3 years ago
Faunal succession and fossil correlation relate to the principle of superposition because there are fossils of certain organisms
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Answer:

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3 years ago
If 700 g of water at 90 °C loses 27 kJ of heat, what is its final temperature?​
Phoenix [80]

Answer:

If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial

In this case:

  • Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)
  • c=4.186 \frac{J}{g* C}
  • m=700 g
  • ΔT= Tfinal - Tinitial= Tfinal - 90 °C

Replacing:

27,000 J=4.186 \frac{J}{g* C}*400 g* (Tfinal - 90C)\\

Solving:

27,000 J=1,674.4 \frac{J}{C}* (Tfinal - 90C)

\frac{27,000 J}{1,674.4 \frac{J}{C}} =(Tfinal - 90C)

16.125 °C= Tfinal - 90 °C

Tfinal= 16.125 °C + 90 °C

Tfinal= 106.125 °C

<u><em>If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C</em></u>

3 0
3 years ago
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