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Sauron [17]
2 years ago
5

Cl2 reacts with the element Sr to form an ionic compound. Based on periodic properties, identify a molecule, X2, that is likely

to to react with Sr in a way similar to how Cl2 reacts with Sr. Justify your choice.
Chemistry
1 answer:
Leto [7]2 years ago
3 0

Answer:

The bromine molecule, Br2 will likely react with Sr in a similar way to chlorine molecule, Cl2.

Explanation:

Chlorine belongs to group 7 of the periodic table. Elements in this group are known as halogens. Its molecule has the formula Cl2. The next element in the group after chlorine is bromine. Its molecule has a formula Br2. It has similar properties as chlorine, therefore, it would react with Strontium in a similar way to chlorine.

Elements belonging to the same group of the periodic the table have similar chemical properties as they contain the same number of valence electrons. Chlorine and Bromine both belong to group 7 of the periodic table and each have seven valence electrons. They both react with metals to form salts even though reaction with chlorine is more vigorous as it is more reactive than bromine. They both form negatively-charged ions with a charge of -1.

Reaction of Strontium with the halogens:

With chlorine: Sr + Cl2 ---> SrCl2

With bromine: Sr + Br2 ---> SrBr2

The salts formed are both crystalline salts with high melting and boiling points.

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Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow  Na_2SO_4(aq) + CO_2(g) + H_2O(l)

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

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