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3 years ago

Which chemical equation shows that the total mass during a chemical reaction stays the same?

Chemistry

2 answers:

k0ka [10]3 years ago

5 0

Answer:

Option A) Mg + Cl₂ → MgCl₂

Explanation:

The law of conservation of mass is guaranteed in a chemical equation. Since the mass of the atoms do not change, that means that the number of each kind of atoms in the reactant side is equal to the number of atoms of the same kind in the product side.

The first equation is:

A) Mg + Cl₂ → MgCl₂

Number of atoms:

atom Reactant side Product side

Mg 1 1

Cl 2 2

Therefore, the table displays that there are the same number of atoms of each kind on both sides, showing that the total mass during the chemical reaction stays the same.

B) NaOH + MgCl₂ → NaCl + MgOH

This equation displays 2 atoms of Cl on the left side and 1 atom of Cl on the right side; thus, it is not showing that the total mass stays the same during the chemical reaction.

C) 2Na + 2H₂O → NaOH + H₂

Neither the sodium, nor oxygen, nor hydrogen atoms are balanced. Thus, this does not show that the total mass stays the same.

D) H₂O + O₂ → H₂O

The reactant side contains 3 oxygen atoms and the product side contains 1 atoms of oxygen; thus, this is not balanced: it does not show that the total mass stays de same during the chemical reaction.

Triss [41]3 years ago

4 0

Answer:

Explanation:

Mg + Cl2 → MgCl2 obeys the law of conservation of mass since the same number of each type of atom appears on both sides of the reaction.

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3 years ago

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3 years ago

Read 2 more answers

1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?

love history [14]

Answer:

$m=1.325gNa_2CO_3$

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

$eq-g=0.5eq-g/L*0.050L=0.025eq-g$

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

$mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3$

Finally, the mass is computed by using its molar mass (106 g/mol)

$m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3$

Regards.

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3 years ago