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jasenka [17]
3 years ago
14

Which orbital is partially filled in the Zirconium (ZI) atom?

Chemistry
1 answer:
Nikolay [14]3 years ago
3 0

Answer:

4d orbital.

Explanation:

Hello!

In this case, since zirconium's atomic number is 40, we fill in the electron configuration up to 40 as shown below:

1s^2,2s^2,2p^6, 3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^2

Thus, the orbital 4d is partially filled.

Best regards!

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in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rat
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Answer:

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

ln k= -(Ea/R) (1/T) + ln A

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

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a)

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

c)

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

8 0
3 years ago
If there is currently 50kg of U-235 present in Oklo, how much must have been present 750 million years ago when the reaction too
Viefleur [7K]
To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.

The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:

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Now, take this scale factor and multiply it by the current mass, like this:

50 \times 1.066 = 53.3
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:

50 + 53.3 = 103.3
I hope this helped you. Fell free to ask any further questions.
7 0
3 years ago
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Two non metal combined together and form the bond is called covalent bond
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8 0
2 years ago
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White raven [17]
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