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3 years ago

What is the ratio of Al 3+ ions to S 2- ions in a neutral compound?

2 answers:

4 0

In a neutral ionic compound, you can determine its sub-scripts by simply flipping the ionic charges and dropping the signs: so AlS would be Al2S3

mr_godi [17]3 years ago

4 0

Answer:

The ratio of aluminium to sulfur is 2:3

Explanation:

The ratio of the elements of the neutral compound can be computed from the ionic charges that makes up the compound. The element that makes up the compound is aluminium and sulfur. From the ions aluminium is the cation with a charge of 3+ . The aluminium ion is the cations because it loses 3 electrons from the bonding between it and sulfur. The sulfur is the anion as it gains electrons from the cations.

Aluminium has a charge of 3+ and sulfur has a charge of 2- . The atom of element that makes up the neutral compound can be computed when you cross multiply the charges.

Al3+ and S2-

cross multiply the charges

Al2S3.

The ratio of aluminium to sulfur is 2:3

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pantera1 [17]

Answer:

2.2 moles

Explanation:

n = CV

n = 0.4×5.50

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2 years ago

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What would be the effect on the nucleus of an atom if it emitted one alpha particle and one gamma ray

Oksana_A [137]

Unstable, radioactive atom emits alpha and gamma ray to transform into a stable atom.

Emitting an alpha particle, decreases the mass number by four and atomic number by two.
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3 years ago

When water evaporates, what happens to the molecules that make up the liquid water?

Rama09 [41]

Answer:

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Explanation:

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3 years ago

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Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

<u>Answer:</u> The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$