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Bingel [31]
3 years ago
13

What are the two different ions present in the compound al(no3)3?

Chemistry
1 answer:
Aleks [24]3 years ago
5 0

Al^{3+} and NO_3^{-}

Aluminum nitrate is an ionic compound, meaning that it is made up cations and anions. By convention when writing the formula for an ionic compounds the cation part shall be placed in front the anions. Thus the cation would contain aluminum and the anion shall be a polyatomic nitrate ion consisting of nitrogen and oxygen. The bracket with subscript 3 attached resembles the ratio between the two species, aluminum ions and nitrate ions. For each aluminum ion in the compound there shall be three corresponding nitrate ions.

Aluminum form ions of charge +3 whereas nitrate ions are of charge -1. Write the charge of the ion as the superscript in reverse (i.e., write 3+ as in Al^{3+} and - in NO_3^{-}, omitting any 1.)

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7. Ionic bonds are at one end of the bond spectrum. What kinds of bonds are at the other end?
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Answer:

non-polar covalent bonds

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4 0
2 years ago
1. What is the wavelength of a photon of light with a frequency of 6.10 x 10^15 Hz?
Blababa [14]

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6 0
3 years ago
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IrinaVladis [17]
There's 18 atoms in ammonium phosphate
5 0
3 years ago
Brainliest will be given to correct answer :)
Lady_Fox [76]

Answer: \Delta G for the reaction   is -90kJ

Explanation:

The balanced chemical reaction is,

2H_2S(g)+SO_2(g)\rightarrow 3S_{rhombic}(s)+2H_2O(g)

The expression for Gibbs free energy change is,

\Delta G=[n\times G_{products}]-[n\times G_{reactants}]

Putting the values we get :

\Delta G=[3\times G_f{S,rhombic}+2\times G_f{H_2O}]-[2\times G_f{H_2S}+1\times G_f{SO_2}]

\Delta G=[(3\times 0kJ/mol)+(2\times -229kJ/mol)]-[(2\times -34kJ/mol) +(1\times -300kJ/mol)]

\Delta G=-90kJ

Thus \Delta G for the reaction2H_2S(g)+SO_2(g)\rightarrow 3S_{rhombic}(s)+2H_2O(g)   is -90kJ

5 0
3 years ago
How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o
natka813 [3]
The balanced chemical reaction is written as:

2K + F2 = 2KF

We are given the amount of potassium to be used in the reaction. THis will be the starting point. We do as follows:

23.5 g K ( 1 mol / 39.10 g) (1 mol F2 / 2 mol K ) ( 22.4 L / 1 mol ) = 6.73 L F2 gas needed
7 0
3 years ago
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