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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Answer:
B on edge
Explanation:
the piles of the magnetic field generated around the armature are attracted to the opposite poles of the permanent magnent. As the opposite poles align, the commutator reverses the current direction so like poles are aligned and the armature continues to spin
Answer:
The air resistance ,F(air) = 294N
Explanation:
Using Nweton's 2nd Law. The sum of the vertical forces must be equal to zero because when terminal speed is reached,the acceleration becomes zero.
EFvertical = mg - F(air) = ma
F(air) = mg = 30kg × 9.8m/s^2
F(air) = 294N
Answer:
what do u need help on
I went to ur profile and I don't know what I'm looking for
I'm going to answer this by using rounded numbers for the atomic masses. You need only go back and put the numbers in from your periodic table. My answers will be close, but not what you should get.
Find the Molar Mass of MgCl2
Mg = 24 grams
2Cl = 2 * 35.5 = 71 grams
Total = 95 grams
Find the mols in 339 grams
1 mole = 95 grams
x mol = 339
Solve
339 = 95x Divide by 95
339/95 = x
x = 3.67 mols Answer