Answer:
Most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.
Explanation:
- In a period as we advance from left to right, the number of valence electrons in the same shell increases due to which the effective nuclear charge increases and thus the atomic size decreases.
- In d-block atomic radius initially decreases then remains constant and increases towards the end.
- As one moves from Sc (scandium) to Zn (zinc), the effective nuclear charge increases by a factor of 1, this is because:
- The number of electrons are low in the inner shell.
- The shielding power of d-orbital is low.
- Inter electronic repulsions will be operating at a value less than the nuclear charge, which will result in decrease in atomic radii.
- As the number of electrons in the inner orbital increases the outer electrons repel one another which enables them to push away.
- Although d-orbital has less shielding power, the number of electrons present in it are high. Hence, the electron-electron repulsive force becomes dominant, this results in an increase in the atomic radii.
Therefore, most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.
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The correct option is A. the isobaric process represented by the PV diagram
<h3>What is isobaric process?</h3>
- An Isobaric method is a thermodynamic revolution taking place at consistent pressure. The period isobaric has been derived from the Greek words “iso” and “baros” indicating equal intimidation.
- As such, the continued pressure is obtained when the importance is expanded or acquired. This basically neutralizes any pressure change due to the transfer of heat.
- In an isobaric procedure, when the heat is transferred to the system some work is done. Nevertheless, there is even a change in the internal energy of the system.
- This additionally means that no amounts as in the first law of thermodynamics evolve zero.
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Answer:
V = 49.05 [m/s]
Explanation:
We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.
where:
Yo = initial position = 0
y = final position [m]
Vo = initial velocity = 0
t = time = 5 [s]
g = gravity aceleration = 9.81 [m/s^2]
The initial speed is zero, as the body drops without imparting an initial speed. Therefore:
y = 0 + (0*5) + (0.5*9.81*5^2)
y = 122.625[m]
Now using the following equation we can find the speed it reaches during the 5 seconds.
![v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f}=\sqrt{2*9.81*122.625} \\v_{f}=49.05 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bi%7D%20%5E%7B2%7D%2B%282%2Ag%2Ay%29%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B2%2A9.81%2A122.625%7D%20%5C%5Cv_%7Bf%7D%3D49.05%20%5Bm%2Fs%5D)
