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svetlana [45]
2 years ago
8

Young fronds are called____ because of how they are coiled

Chemistry
2 answers:
topjm [15]2 years ago
4 0

Answer:

Fiddlehead

Explanation:

Young fronds are called Fiddleheads

because of how they are coiled

likoan [24]2 years ago
3 0

Answer:

Fiddleheads or koru

Explanation:

The coiled, immature leaves are called fiddleheads. Young fronds, called fiddleheads because of their striking resemblance to the head of a violin, start out tightly curled at the base of the root. New, or young fronds are produced from the rhizome. They are tightly coiled into a spiral (called a fiddlehead or koru), and these slowly uncoil as they mature.

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In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions)
jeyben [28]

Answer:

Li2S> Na2S> K2S> CsS

Explanation:

The lattice energy of ionic species depends on the relative sizes of ions in the ionic compounds. As the size of ions increases, the lattice energy decreases and vice versa.

When the size of the anions are the same, the lattice energy now depends on the relative sizes of the cations. Therefore, since all the compounds are sulphides and the order of magnitude of ionic sizes is: Li^+ < Na^+ < K^+ < Cs^+.

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3 years ago
The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

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123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

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