Answer:
Al(NO3)3(s)--------> Al^3+(aq) + 3NO3^-(aq)
Explanation:
The equation shown above describes the dissolution of Al(NO3)3 in water using the lowest coefficients.
This occurs when solid Al(NO3)3 is added to water. It dissolves to give rise to ions as shown. This is a property of all ionic substances.
it is a molecule* that can be joined with other molecules that are identical to form a polymer*
key words :
a molecule:
a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
a polymer:
a substance that has a molecular structure consisting chiefly or entirely of a large number of similar units bonded together
hope this helped, good luck in future studies !
-A
Above question is incomplete. Complete question is attached below
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Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case,
reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is
least reactive.
On other hand,
reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is
most reactive.
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
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu