High School

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Predict the products and write each balanced chemical equation.

b. Na\(_2\)CO\(_3\) + Sr(NO\(_2\))\(_2\) →

A. 2NaNO\(_2\) + SrCO\(_3\)

B. Na\(_2\)CO\(_3\) + 2Sr(NO\(_2\))\(_2\) → 2NaNO\(_2\) + SrCO\(_3\)

C. Na\(_2\)CO\(_3\) + Sr(NO\(_2\))\(_2\) → NaNO\(_2\) + SrCO\(_3\)

D. 2Na\(_2\)CO\(_3\) + Sr(NO\(_2\))\(_2\) → 2NaNO\(_2\) + SrCO\(_3\)

Answer :

Final answer:

The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and strontium nitrite (Sr(NO2)2) is a) Na2CO3 + Sr(NO2)2 → 2NaNO2 + SrCO3,

Explanation:

The products and write a balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and strontium nitrite (Sr(NO2)2).

For this reaction, a double-replacement reaction is expected, and the products would be sodium nitrite (NaNO2) and strontium carbonate (SrCO3).

The balanced chemical equation for this reaction is:
Na2CO3 + Sr(NO2)2 → 2NaNO2 + SrCO3

This equation balances the atoms of each element on both the reactant and product sides, satisfying the Law of Conservation of Mass.

Therefore, the correct answer is a) Na2CO3 + Sr(NO2)2 →2NaNO2 + SrCO3

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