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2 years ago

An isotope will lose nucleons, protons, or neutrons in order to become more

Chemistry

1 answer:

Ne4ueva [31]2 years ago

3 0

An isotope will lose neutrons in order to become more stable.

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Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to:

$5$ $\rm O_2$ molecules.

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

5 0

2 years ago

If copper carbonate system is heated for too short a time, what will you notice for mass percent of copper be too large or too s

sammy [17]

Answer:

The mass percent of copper as element is the same.

Explanation:

First of all we need the reaction that is presented below:

$CuCO_{3 (s)} +Heat$ → $CuO_{(s)} + CO_{2 (g)}$

The mass percent of copper (Cu) as element is the same because of during the reaction the element only transform its nature from copper carbonate ( $CuCO_{3 (s)}$ ) to copper oxide ( $CuO_{(s)}$ ), the latter is a solid and will remain in the system.

On the other hand, you will note that the global percentage mass will be small because of the reaction produce ( $CO_{2 (g)}$ ) that is a gas and this one will escape for the system.